. .v'.'VfI^vl^i
'^:n?«^?;,^ V <^'vf ?Vr^^:^^*i'
.'■ ■••»■:. "(
•
€
CO
e^^
1
jP
Wi
»— *■
Cjj
3
J
rH
8
CQ
<D
rH
rH
»4
1
1
f^^^"'
J.
1
1
GEOMETRICAL DRAWING AND DESIGN.
0m,
MACMILLAN AND CO., Limited
LONDON • BOMBAY • CALCUTTA
MELBOURNE
THE MACMILLAN COMPANY
NEW YORK • BOSTON • CHICAGO
DALLAS • SAN FRAKCLSCO
THE MACMILLAN CO. OF CANADA, Ltd,
TORONTO
GEOMETRICAL
DRAWING AND DESIGN
BY
J. HUMPHREY SPANTON
GOLD MEDALLIST, ROYAL ACADEMY OF ARTS, LONDON
INSTRUCTOR IN DRAWING TO NAVAL CADETS IN H.M.S. "hRITANNIa'
AUTHOR OF "complete PERSPECTIVE COURSE*'
ADAPTED TO THE REQUIREMENTS OF
THE BOARD OF EDUCATION
MAC^riLLAN AND CO., LIMITED
ST. MARTIN'S STREET, LONDON
1913
'S'SIOO
1
COPYRIGHT.
First Eflition 100?.
Repiintcd ]004; with alterations and additions 1006.
Reprinted 1009, 1011, with additions 1013.
Me.
Si
GLASGOW : PRINTKD AT THE UNIVERSITV PRESS
BY ROBERT IMACt.EHOSE AND CO. LTD.
■Mi'
PREFACE.
A COURSE of geometrical drawing or practical geometry pro
vides a valuable preliminary training for so many handicrafts
and professions that it must be regarded as essential to all
students whose work is to be adapted to modern requirements.
The Engineer, the Architect, the Soldier, and the Statistician,
all have recourse to the assistance of practical geometry to solve
their problems or to explain their methods. Every day the
graphic treatment of subjects is finding its application in new
directions ; and to be able to delineate the proportions of any
scheme places in the hand an invaluable tool for the execution
of work of practical value.
The Author's complete geometrical course has now for some
time been widely used for the advanced parts of the subject;
such as Projection, Section and Interpenetration of Solids.
Hence the publication of its simpler parts as an introduction
to Design would seem likely to meet with favour.
The geometrical basis of ornamentation is the rational one,
though youth and fancy might condemn it as a chaining of
Pegasus and the curbing of imagination. It is no doubt the
height of art to conceal the scheme of treatment and delight
the eye with novel suggestions of development. But underlying
all is the demand of Nature for order and rhythm, such as
can be obtained by a study of geometrical figures and designs.
The course of work prescribed in Geometrical Drawing (Art)
by the Board of Education aims at giving students the ability
to construct ordinary geometrical figures, and the power to
PREFACE.
apply these figures as the basis of ornamental and decorative
work. In the preparation of the present volume these intentions
have been borne in mind, but the scope of the work has not
been limited by the syllabus of the subject. The greater
part of the book contains an elementary course of constructive
geometry suitable for all students, and the sections which show
the applications of geometrical constructions to design, while
of especial importance to students of appHed art, are not
without interest to students of science. Moreover, familiarity
with the constructions in the early part of the book provides
the best introduction a pupil could have to the study of formal
geometry.
The Editor's thanks are due to Prof Thos. C. Simmonds,
the Headmaster of the Municipal School of Art, Derby, for
his valuable advice, and to Mr. E. E. Clark, his assistant, by
whom the drawings for the sections on Design have been
furnished. Acknowledgment must also be made of assistance
rendered by Prof R. A Gregory in arranging the book and
seeing it through the press.
CONTENTS.
PAGE
Drawing Instruments and Materials.   . .  i
General Directions,  3
CHAPTER I.
Geometrical Definitions and General Properties, ... 6
PLANE GEOMETRY.
CHAPTER II.
Lines, Triangles, Quadrilaterals, Convergent Lines, and Circles, 15
CHAPTER III.
General Information concerning Polygons,     30
General Methods for constructing Polygons,     32
CHAPTER IV.
Inscribed and Described Figures, 40
CHAPTER V.
Foiled Figures, 53
viii CONTENTS
CHAPTER VI.
PAGE
Tangents a ' tangential Arcs, 57
CHAPTER VIL
Proportional Lines,    72
CHAPTER Vni.
Construction of Plain Scales,        81
,, ,, Comparative Scales, 84
,, ,, Diagonal Scales, 86
,, ,, Proportional Scales, 89
CHAPTER IX.
The Use of the Protractor,  92
The Use of the Sector,   94
CHAPTER X.
Construction of Similar Figures, 101
Principles of Similitude, or the Enlargement and Reduction of
Plane Figures, 103
CHAPTER XL
The Conic Sections — 107
Practical Method for drawing the Conic Sections,   107
Mechanical Methods for drawing the Ellipse,    108
The Ellipse,   108
The Parabola, 112
The Hyperbola,         113
Mechanical Methods for drawing the Parabola and Hyper
bola, 114
Cycloidal Curves, 116
Spirals,          121
CONTENTS ix
SOLID GEOMETRY.
CHAPTER XII.
Page
itroduction, 126
CHAPTER XIII.
Simple Solids in given Positions to Scale,     132
The Regular Solids, 142
Octagonal Pyramids, Cones, Cylinders, and Spheres,   146
CHAPTER XIV.
\ rthographic Projection,    157
,, Lines, 158
,, Planes, 162
CHAPTER XV.
Sections of Solids, Construction of Sectional Areas,   169
,, ,, a Cone,         '75
„ ,, a Cylinder,   I79
,, ,, a Sphere, 180
DESIGN.
CHAPTER XVI.
Construction Lines on which Patterns are arranged,    184
Units of Pattern, . .  193
Spacing of Walls and other Surfaces, 196
Bands and Borders,         197
Defined areas,  202
CONTENTS
CHAPTER XVII.
Lettering
Shields,
Diaper, Chequer, Spot, Powder, ... 
CHAPTER XVIIL
Arch Forms and Tracery, . . . = =
Greek and Roman Mouldings,     =
Gothic Mouldings,   .....
Gothic Piers,  
Miscellaneous Exercises,
Examination Papers in Geometrical Drawing,
PART I.
PRACTICAL PLANE GEOMETRY.
INTRODUCTION.
Drawing Instruments and Materials.
The Drawingboard. — A very convenient size to use for
ordinary purposes is half Imperial (23'' x 16") ; it should be
made of wellseasoned yellow pine, with the edges true and at
right angles to each other.
The Teesquare.— This is a ruler with a cross piece or stock
at the end : it is like the letter T in shape, hence its name.
The blade should be screwed on to the stock, and not mortised
into it, so as to allow of the setsquares being used up to the
extreme margin of the paper, as illustrated in Fig. i. By
keeping the stock of the teesquare pressed closely against the
edge of the drawingboard, we are enabled to draw lines parallel
to each other.
Setsquares. — These are rightangled triangles made with
given fixed angles out of thin pieces of wood or ebonite : the
latter is preferable, as it is not liable to warp. The most useful
angles are those of 45° and 60°.
French Curves. — These are thin pieces of wood cut into a
variety of curves. They are used for drawing fair curves, that
are not arcs of circles, through a succession of points : the
cycloidal curves, for instance.
Scale. — A plain scale about 6 inches long, divided into inches,
with subdivisions of eighths on one edge and tenths on the
other.
CI A
2 geomp:trical drawing and design.
Pencils. — Two degrees of hardness should be used : HH for
drawing in the construction, and F for drawing in the result
with a firmer line.
Drawingpaper.— This should have a hard smooth surface.
Whatman's "hotpressed" is the best for fine work ; but if the
drawing has to be coloured, a damp sponge should first be
drawn across the surface, to remove the gloss. Cartridgepaper
of good quality is suitable for ordinary work.
The most convenient size is "Imperial" (30" x 22";, which can
be cut to half, or quarter Imperial, as desired.
Drawingpins. — These should have short fine points, so as
not to make large holes in the drawingboard.
Dividers. — These are used for setting off distances or
dividing lines. There is a special kind made, called "hair
dividers," one leg of which can be adjusted by means of a
spring and screw : these are very useful for dividing lines, etc.
Rulingpen. — This is used for inking in lines, the thickness
of which is regulated by a screw. Some are made in which the
nib that works against the ruler is of an extra thickness of metal :
this is to prevent the nibs from closing when the pen is pressed
against the ruler.
Bowpencil. — This is a small pair of compasses with one leg
constructed to hold a pencil, used for drawing circles and arcs.
Bowpen. — This is a similar instrument to a bowpencil, but
with a ruling pen for one of its legs instead of a pencil ; it is
used for inking in circles and arcs.
Note. — Both the bov/pencil and bowpen should have hinged
legs ; because, when a number of circles are drawn from the
same centre, they are likely to make a large hole in the paper,
unless the leg used for the centre is kept perpendicular to the
paper. It is also necessary to have the penleg as upright as
possible, otherwise it has a tendency to draw uneven lines.
Compasses.— Fullsized compasses are suppHed, with inter
changeable arms, divider, pen and pencil, opening to 12 inches.
Indian ink should be used for inking in a drawing. It has
several advantages over common ink : it dries quickly ; it does
not corrode the rulingpen ; and the lines can be coloured over
without their running, if a waterproof quality is used.
The most convenient is the liquid Indian ink, sold in bottles,
PRACTICAL PLANE GEOMETRY
as it is always ready for use. The rulingpen should be filled
with Indian ink by means of an ordinary steel nib. If the cake
Indian ink is used, after rubbing it in a saucer, a piece of thin
whalebone should be used for filling the rulingpen.
General Directions.
Keep all instruments perfectly clean : do not leave ink to
dry in the ruHngpen.
In using dividers avoid, as much as possible, making holes
through the paper.
The paper should be firmly fixed to the drawingboard by
a drawingpin at each corner, well pressed down. Do not stick
pins in the middle of the board, because the points of the
dividers are liable to slip into them and make unsightly holes
in the paper.
A pencil sharpened to what is called a " chiselpoint " is
generally used for drawing lines ; it has the advantage of retain
ing its point longer, but a nicelypointed pencil is better for
neat work, as it enables you to see the commencement and
termination of a line more easily.
Fig. I.
Always rule a line from left to right, and slope the pencil
slightly towards the direction in which it is moving ; if this is
done, there is less chance of indenting the paper, which should
always be avoided.
4 GEOMETRICAL DRAWING AND DESIGN.
Having determined the extent of a line, always rub out the
superfluous length ; this will prevent unnecessary complication.
Avoid using Indiarubber more than is necessary, as it tends
to injure the surface of the paper. After inkmg in a drawings
use stale bread in preference to Indiarubber for cleaning it up.
The teesquare should be used for drawing horizontal lines
only (Fig. i) ; the perpendicular lines should be drawn by
the setsquares. If this is done, it is immaterial whether the
edges of the drawingboard are at right angles, because it will
only be necessary to use one of its edges.
For drawing parallel lines that are neither horizontal nor
perpendicular, hold one setsquare firmly pressed upon the
paper and slide the other along its edge (Fig. 2). Geometrical
drawing can be greatly facilitated by the proper use of set
squares, so it is advisable to practise their use.
Fig. 2.
When a problem contains many arcs of circles, it is advisable
to connect each arc with its corresponding centre. Enclose the
centre in a small circle ; draw a dotted line to the arc, termi
nated by an arrowhead (Fig. 3).
In drawing intersecting arcs for bisecting lines, etc., the arcs
should not intersect each other too obtusely or too acutely : the
nearer the angle between the arcs approaches 90° the easier it
will be to ascertain the exact point required.
PRACTICAL TLANE GEOMETRY.
In joining two points by a line, first place the point of the
pencil on one point, then place the edge of the ruler against it,
and adjust the ruler till its edge coincides with the other point.
All the problems should be drawn larger than shown.
Great accuracy is required in drawing the various problems.
Every effort should be made to ensure neatness and precision
in the work.
All arcs should be inked in first, as it is easier to join a hne to
an arc than an arc to a line.
CHAPTER I.
GEOMETRICAL DEFINITIONS,
A point simply marks a position ; it is supposed to have no
magnitude.
A line has length without breadth or thickness. The ex
tremities and intersections of lines are points. A straight line
or right line is one that is in the same direction throughout
its length, and is the shortest that can be drawn between two
points. To produce a line is to lengthen it.
A plane is a flat even surface ; it has length and breadth
only. The intersections of planes are straight lines.
Parallel lines are straight lines in the
same plane, and at equal distances apart
throughout their entire length ; if pro "
duced they would never meet (Fig. 4), ^'^ 4
A circle is a plane figure bounded by a curved line, such that
all straight lines drawn to it from a
certain point are equal. This point is
called the centre, and the curved line
is called the circumference of the circle.
A straight line drawn from the centre
to the circumference, as ce or cd (Fig. 5)
is called a radius. A straight line drawn
through the centre, and terminated at
both ends by the circumference, as ab^
is called a diameter. A semicircle is
half a circle, as adb. A quadrant is a quarter of a circle, as adc.
GEOMETRICAL DEFINITIONS.
An arc is any portion of ihe circumference of a circle, as abc
(Fig. 6). A chord is a straight line join
ing the extremities of an arc, as ac. A
segment is the space enclosed by the
arc and its chord, as f. A sector is the
space enclosed by two radii and the
arc between them, as g. A tangent is
line touching the circumference, as (ie ;
it is always at right angles to the radius
of the circle at the point of contact.
Fig. 9.
An ordinate is a line drawn from a point m
a curve perpendicular to the diameter, as dotted
line in Fig. 7.
An abscissa is the part of the diameter cut
off by the ordinate, as dotted line in Fig. 8.
An angle is the inclination of two
straight lines meeting in a point. This
point is called the vertex of the angle,
as a (Fig. 9). The angle here shown
would be called either dac or cad.
If two adjacent angles made by two straight lines at the
point where they meet be equal, as f^ca and deb (Fig. 5), each
of these angles is called a right angle, and either of the straight
lines may be said to be perpendicular to the other.
A right angle is supposed to be divided into 90 equal parts,
each of which is called a degree. A degree is expressed in
writing by a small circle placed over the last figure of the
numerals denoting the number of degrees — thus 36° means
thirtysix degrees.
The circumference of a circle is supposed to be divided into
360 equal arcs, each of which subtends an angle of 1° at the
centre. Sometimes this arc is itself loosely termed a degree.
An angle containing more than 90° is called an obtuse angle,
as ecd (Fig. 5) ; while an angle containing less than 90° is called
an acute angle, as ace.
8
GEOMETRICAL DRAWING AND DESIGN.
A line is said to be perpendicular to a plane when it is at
right angles to any straight line in that plane which meets it.
Concentric circles have the same centre.
Triangles.
Triangies are closed figures contained by three straight lines.
A triangle which has all its sides equal is
called equilateral (Fig. lo).
N.B. — Such a triangle will always have its
three angles equal, and therefore will also be
equiangular.
A triangle which has two sides (and there
fore two angles) equal is called isosceles (Fig. 1 1)
{isos^ equal ; skelos, a leg).
A scalene triangle has none of its sides equal.
A triangle which has a right angle is called
rightangled (Fig. 12). The side opposite to the
right angle, as ab, is called the hypotenuse {hypo,
under ; te?iein, to stretch).
A triangle which has an obtuse angle
is called obtuseangled (Fig. 1 3).
A triangle which has three acute angles
is called acute angled (Fig. 14).
Fig. 14.
The base of a triangle is its lowest side, as al? (Fig. 10).
The vertex is the point opposite the base, as c (Fig. 10).
The altitude or perpendicular height is a line drawn from
the vertex at right angles to its base, as cd {¥\^. 10).
The median is a line drawn from the veitex to the middle
point of the base.
GEOMETRICAL DEFINITIONS.
Quadrilateral Figures.
Quadrilateral figures are such as are bounded by four straight
lines.
A quadrilateral figure whose
opposite sides are parallel is
called a parallelogram (Fig. 15).
N.B. — The opposite sides and
angles of parallelograms are ^ ^d
equal. fig ^s
A parallelogram whose angles are
right angles is called a rectangle
(Fig. 16) or oblong.
A rectangle which has its sides equal
called a square (Fig. 1 7).
A parallelogram whose angles
are not right angles is called
a rhombus, if its sides are all
equal (Fig. 18), or a rhomboid if
the opposite sides alone are
equal — Gk. rhombos^ from rhein
bein^ to twirl, from some likeness
to a spindle.
All other quadrilaterals are called trapeziums.
A line joining two opposite angles of a quadrilateral figure is
called a diagonal, as the dotted line ab (Fig. 1 5).
Polygons.
A polygon is a plane figure which has more than four angles.
A polygon which is both equilateral and equiangular is called
regular.
10 GEOMETRICAL DRAWING AND DESIGN.
A polygon of five sides is called a pentagon.
„ „ six „ „ hexagon.
„ „ seven „ „ heptagon.
„ „ eight „ „ an octagon, etc.
Solids.
A solid has length, breadth, and thickness. A solid bounded
wholly by planes is called a polyhedron {poly^ many ; hedra^ a
side).
A solid bounded by six planes or faces, whereof the opposite
ones are parallel, is called a parallelepiped {pa?aUelos, parallel ;
and epipedon^ a plane).
A parallelepiped whose angles are all right angles is called a
rectangular parallelepiped or orthohedron {ort/ios, right ; and hedra^
a side).
An orthohedron with six equal faces is called a cube {kudos, a
die).
A polyhedron, all but one of whose faces meet in a point, is
called a pyramid (Gk. pyrainis, a pyramid).
Pyramids are often named, after the shape of their bases,
triangular, square, etc.
A polyhedron, all but two of whose faces are parallel to one
straight line, is called a prism (Gk. prisma, irom. prizein, to saw,
a portion sawn off).
If the ends of a prism are at right angles to the straight line
to which the other faces are parallel it is called a right prism.
Prisms are often named, after the shape of their ends,
triangular, hexagonal, etc.
A cylinder is a solid described by the revolution of a rect
angle about one of its sides which remains fixed. This fixed
line is called the axis of the cylinder.
A right circular cone — generally spoken of simply as a cone
— is a solid described by the revolution of a rightangled triangle
about one of the sides containing the right angle, which side
remains fixed. This fixed line is called the axis of the cone ;
the base is a circle, and the point opposite the base is called
the vertex.
A solid bounded by a closed surface, such that all straight
GEOMETRICAL DEFINITIONS.
lines drawn to it from a certain point are equal, is called a
sphere (Gk. sphatra, a ball).
The point referred to is called the centre of the sphere.
Technical Definitions.
A plane parallel to the ground, or, more strictly speaking,
parallel to the surface of still water, is called a horizontal plane.
A vertical plane is a plane at right angles to a horizontal
plane.
A horizontal line is a line parallel to a horizontal plane.
A vertical line is a line at right angles to a horizontal plane.
Fig. 19.
The plan of an object is the tracing made on a horizontal plane
by the foot of a vertical line, which moves so as to pass succes
sively through the various points and outlines of the object, as
A and B (Fig. 19).
12
GEOMETRICAL DRAWING AND DESIGN.
An elevation of an object is the tracing made on a vertical
plane by the end of a horizontal line, at right angles to the
vertical plane, which moves so as to pass successively through
the various points and outlines of the object, as A' and B'
(Fig. 19).
N.B. — A is an object parallel to the vertical plane, and B an
object inclined to it.
General Properties of some of the Figures
already described.
If two lines cross each other the opposite angles are always
equal. The angle acb is
equal to the angle dee, and
the angle aed is equal to the
angle bee (Fig. 20). ^"
The two adjacent angles
are equal to two right angles ;
the angles aed and aeb^ for
instance, as well as the angles bee and
Fig. 20.
:d.
Triangles.— The three angles of a triangle con
tain together 180°, or two right angles ; so if two
angles are given, the third angle can always be
found. F^or example, if one angle is 70° and the
other 30°, the remaining angle must be 80° (Fig. 21).
70° + 30°= 100°.
1 80° 1 00° = 80°.
The exterior angle of a
triangle is equal to the two
opposite interior angles.
The angle abe is equal to the
angles bed and edb together ;
in the same way the angle
ede is equal to the two angles ^
bed and ebd (Fig, 22).
b d e
Fig. 22.
If we multiply the base by half the altitude, we get the area
of a triangle ; or half its base by its altitude will give us the
same result.
GEOMETRICAL DEFINITIONS.
!3
Triangles of equal bases drawn between parallel lines are
equal in area, and lines drawn parallel to their bases at equal
heights are equal in length, as the dotted lines shown (Fig. 23).
Fig. 23.
If we bisect two sides of a triangle and join
the points of bisection, we get a line that is
always parallel to the third side (Fig. 24).
Quadrilaterals. — If we bisect the four
sides of a quadrilateral figure and join
the points, it will always give us a
parallelogram, as shown by dotted
lines. The reason for this will be
apparent by the principle shown in the
preceding figure if we draw a diagonal
of the quadrilateral, so as to form two
triangles (Fig. 25).
Fig. 25.
Parallelograms drawn between parallel lines on equal bases
are always equal in area, and parallel lines drawn at equal
heights are always equal to each other and to the bases, as
shown by dotted lines (Fig. 26).
Fig, 26.
14
GEOMETRICAL DRAWING AND DESIGN.
Semicircles. — Any two lines drawn
from the extremities of the diameter,
to a point on the circumference of a
semicircle, will form a right angle.
Fig. 27.
EXERCISES.
Note. — Feet are represented by one dash {'), and inches by two
dashes (") ; 3 feet 6 inches would be written thus — 3' 6".
1. Draw lines of the following lengths: 3", 45", 2", ig", 2.25",
3.50", 1.75".
2. Draw an acute angle, and an obtuse angle.
3. Draw the following triangles, viz. equilateral, scalene, isosceles,
obtuseangled, rightangled, and acuteangled.
4. Draw a rightangled triangle, and write the following names to
its different parts, viz. hypotenuse, vertex, base, median, and altitude.
5. Draw the following figures, viz. rectangle, rhombus, square,
rhomboid, trapezium, and parallelogram.
6. Draw a circle, and name the different parts, viz. sector, radius,
chord, arc, diameter, segment, and tangent.
CHAPTER 11.
PROBLEMS ON LINES, TRIANGLES, QUADRILATERALS,
CONVERGENT LINES, AND CIRCLES.
Lines.
1. To bisect a given straight line
AB.
From A and B as centres, with
any radius greater than half the
hne, describe arcs cutting each
other in C and D ; join CD. The
straight hne CD will bisect AB in
E. Also the line CD will be per
pendicular to AB.
2. To bisect a given arc AB.
Proceed in the same way as in
Problem i, using the extremities
of the arc as centres. The arc
AB is bisected at E.
i6 GEOMETRICAL DRAWING AND DESIGN.
3. To draw a line parallel to a given line AB through a given
point C.
p ^ ^ Take any point D in line
,''p ^ AB, not opposite the point
\ \ C ; with D as centre, and DC
\ V as radius, describe an arc
\ « cutting AB in E, and from
\ 15 £_'. C as centre, with the same
c_ ' ■" B radius, draw another arc DF ;
Fig. 30 set off the length EC on DF ;
a line drawn through CF will be parallel to AB.
4. To draw a line parallel to a given line AB at a given distance
from it.
p r. Let the length of the line C
"^ '' "" represent the given distance.
Take any points D and E
in line AB as centres, and
with C as radius describe
Ad E ^ arcs as shown ; draw the line
FG as a tangent to these
arcs. FG will be parallel to
Fig. 31 AB.
5. From a point C in a given line AB, to draw a line
perpendicular to AB.
F
c
Fig. 32.
At the point C, with any
radius, describe arcs cutting
AB in D and E ; with D
and E as centres, and with
any radius, draw arcs inter
secting at F ; join FC,
c < which will be perpendicular
— ! r, to AB.
PROBLEMS.
6. To draw a perpendicular to AB from a point at, or near, the end
of the given line.
With B as centre, and with
any radius BC, draw the arc
CDE ; and with the same
radius, starting at C, set off
points D and E ; with each of
these two points as centres
and with any radius, draw arcs
cutting each other at F ; join
FB, which will be perpendicular
to AB.
F^
Fig 33
7. To draw a line perpendicular to a given line, from a point
which is without the line.
C
Let AB be the given line
and C the point.
With C as centre, and with
any radius greater than CD,
draw arcs cutting the line i— ___
AB in E and F ; from these ^ ^^ ~
points as centres, with any
radius, describe arcs inter
secting at G ; join CG, which
will be perpendicular to AB.
Fig. 34.
8. To draw a perpendicular to AB from a point opposite, or nearly
opposite, to one end of the Une.
r
Let C be the given point.
Take any point D in AB, not
opposite the point C ; join
CD and bisect it in E ; with
E as centre, and EC as radius,
draw the semicircle CFD ;
join CF, which will be per
pendicular to AB.
zy
Fig. 35
GEOMETRICAL DRAWING AND DESIGN.
To divide a given line AB into any number of equal parts.
Take five for example.
.D
B
Fig. 36.
into five equal parts.
F'rom one extremity A,
draw AD, at any angle to
AB ; and from B, draw BC,
parallel to AD. With any
convenient radius, set off
along AD, commencing at A,
^ , K'" ' four parts (the number of
parts, less one, into which it
is required to divide the given line), repeat
the same operation on BC, commencing at
B, with same radius ; join the points as
shown, and the given line AB will be divided
10.
c
/A
Another Method.
Draw AC at any angle to
AB ; AC may be of any con
venient length. With any
radius, mark off along AC
the number of equal parts re
\ quired ; join the last division
\ C with B ; draw lines from
__i all the other points parallel to
B CB, till they meet AB, which
will be divided as required.
Fig. 37
m. From a given point B, in a given line AB, to construct an
angle equal to a given angle C.
From point C of the given
angle as centre, and with
any radius, draw the arc
EF : and with the same
radius, with B as centre,
draw the arc GH ; take the
length of the arc EF, and
set it off on GH ; draw the
line BD through H. Then
the angle GBH will be equal
to the given angle C.
PROBLEMS.
12.
To bisect a given angle
ABC.
From B as centre, and with
any radius, draw the arc AC ;
from A and C as centres, with
any radius, draw the arcs inter
secting at D ; join DB, which
will bisect the angle ABC.
13. To trisect a right angle.
Let ABC be the right angle.
From B as centre, and with any
radius, draw the arc AC ; with
the same radius, and A and C as
centres, set off points E and D;
join EB and DB, which will
trisect the right angle ABC.
Fig. 40.
14. To trisect any angle ABC^
From B, with any radius describe the arc AHC
the angle ABC ; join A and C
cutting the bisector in D ; with
D as centre, and with DA as
radius, describe the semicircle
AGC, cutting the bisector in G ;
and with the same radius, set
off the points E and F from A
and C ; join AG ; take the
length AG and set it off from
H along the line GB, which
will give the point I ; join EI
and FI, which will give the
points J and K on the arc AHC ;
join J and K with B, which wil
1 This is one of the impossibiUties of geometry ; but this problem, devised
by the author, gives an approximation so near, that the difference is imper
ceptible in ordmary geometrical drawing.
bisect
Fig. 41.
trisect the angle ABC.
GEOMETRICAL DRAWING AND DESIGN.
Triangrles.
D
15. To construct an equilateral
triangle on a given line AB.
From A and B as centres, and
with AB as radius, describe arcs
cutting each other at C ; join C
with A and B. Then ABC will
be an equilateral triangle.
16. On a given base AB to con
struct an isosceles triangle,
the angle at vertex to be equal
to given angle C.
Produce the base AB to E,
and at A construct an angle FAE
making with AE an angle equal
to C. Bisect the angle FAB
by the line AD. From B draw
a line making with AB an angle
equal to DAB, and meeting
AD in D. ADB will be the
isosceles triangle required.
17. On a given base AB, to con
struct an isosceles triangle,
its altitude to be equal to a
given line CD.
Bisect AB at E, and erect
a perpendicular EF equal in
height to the given line CD ;
join AF and BF, then AFB
will be the isosceles triangle
required.
Fig. 44.
PROBLEMS.
21
18. To construct a triangle, the three sides A, B, and C being given.
Make the base DE equal to
the given line A. From D as
centre, and with radius equal to
line B, describe an arc at F,
and from E as centre, and with
radius equal to line C, draw
another arc, cutting the other at
F ; join FD and FE, which will
,give the triangle required.
19. To construct a triangle with
two sides equal to given lines
A and B, and the included
angle equal to C.
Make an angle DEF equal
to given angle C, in required
position. Mark off EF equal
to line A, and ED equal to
line B ; join DF. DEF is the
triangle required.
20. To construct a triangle with
a perpendicular height equal
to AB, and the two sides
forming the vertex equal to
the given lines C and D.
Through B draw the Hne EF
at right angles to AB. From A
as centre, and with radii equal
to the lengths of the lines C
and D respectively, draw arcs
cutting the line EF ; join AE
and AF, which will give the
triangle required.
Fig.
Fig 47
GEOMETRICAL DRAWING AND DESIGN.
A
/
Fig. 48.
^T
Fig. 50.
21. To construct a triangle, on
the given base AB, with
one base angle equal to C,
and the difference of the
sides equal to given line D.
At end A of the base, con
struct an angle equal to the
given angle C. Cut off AF
equal to line D, the given'
difference of the sides ; join
FB. Bisect FB at right
angles by a line meeting AF
produced in E ; join EB.
AEB is the required triangle.
22. To construct a triangle on
a base equal to given line
A, with vertical angle equal
to D, and sum of the two rc
^ maining sides equal to BE.
Draw the Hne BE, and at
E construct an angle making
with BE an angle equal to
half the given angle D.
From point B, with radius
equal to given line A, draw
an arc cutting EC at C ; join
BC. Bisect CE at right
angles by line FG ; join FC.
BCF will then be the tri
angle required.
23. To construct a triangle
with two sides equal to
the given lines A and B
respectively, and the in
cluded median equal to
given line C.
Draw a triangle with the
side DE equal to given line
A, the side DF equal to
PROBLEMS.
2^
given line B, and the third side FE equal to twice the given
median C. Bisect the line FE at G. Join DG and produce i*^
to H, making GH equal to DG ; join EH.
DEH will be the triangle required.
24. On a given base AB to
describe a triangle similar
to a given triangle DEF.
Make angles at A and B
equal respectively to the
angles at D and E. Produce
the lines to meet at C. Then
ABC will
required.
25.
be the triangle
Quadrilaterals.
To construct a square on a given base AB.
C
At point A erect AC perpen
dicular to AB and equal to it.
With B and C as centres, and
radius equal to AB, draw inter
secting arcs meeting at D ; join
DC and DB.
CABD is the square required.
Fig. 52.
26. To construct a square on a given diagonal AB.
c
Bisect AB at right angles by
the line CD. Mark off EC and
ED equal to EA and EB ; join
CA and AD, and CB and BD.
CADB is the square required.
\\E\
kl
l^ig 53
24
GEOMETRICAL DRAWING AND DESIGN.
27. To construct a rectangle with sides equal to given lines A and B.
Draw the line CD equal
to given line A. At D erect
a perpendicular DF equal to
given line B. With C as
centre and radius equal to
line B, and with F as centre
and radius equal to line A,
draw arcs intersecting each
other at E ; join EC and
EF, which will give the rect
angle required.
Fig. 54
28. To construct a rectangle with diagonal equal to given line A,
F^'"~~
^11'' G
and one side equal to
given line B.
Draw the line CD equal
to given line A. Bisect CD
at E. With E as centre, and
with radius EC, describe a
circle ; from C and D as
centres, and with given line
B as radius, set off the points
F and G ; join CO, CD,
DF, and FC.
~ * CGDF is the required
~ rectangle.
Fig. 55
29. To construct a rhombus with sides equal to given line A, and angle
equal to given angle C.
Make the base DE equal
to given line A. At D con
struct an angle equal to given
angle C. Set off DF equal
to DE ; with F and E as
centres, and radius equal to
DE, draw arcs intersecting
at G ; join FG and EG.
DEGF will be the required
rhombus.
B—
Fig. 56.
PROBLEMS.
25
Bisect the angle BEF —'"
30. To draw a line bisecting the angle between two given
converging lines AB and CD, when the angular point is
inaccessible. /\
From any point E in AB,
draw a line EF parallel to
CD.
by the line EG. At any
point H between E and B,
draw HL parallel to EG.
Bisect EG and HL in M and
N. Join MN, which, pro
duced, is the bisecting line C
required.
31. Through the given point A, to draw a line which would, if
produced, meet at the same point as the given lines BC and
DE produced.
l'ig 57
Draw any convenient line
FG; join A^FandAC. Draw
any line HK parallel to FG.
At H draw the line HL
parallel to FA, and at K
draw the line KL parallel to
GA, cutting each other at L.
Draw a line through L and
A ; AU is the convergent
line required.
Circles.
32. To find the centre of a
circle.
Draw any chord AB, and
bisect it by a perpendicular
DE, which will be a diameter
of the circle. Bisect DE in
C, which will be the centre
of the circle.
Fig 59
26
GEOMETRICAL DRAWING AND DESIGN.
33.
To draw a circle through three given points A, B, C.
Join AB and BC. Bisect
AB and BC by perpendi
culars cutting each other at
D, which will be the centre
of the circle. From D as
centre, and DA as radius,
describe a circle, which will
then pass through the given
points A, B, C, as required.
Note.— The two following
problems are constructed in
the same manner.
Fig. 60.
34. To draw the arc of a circle through three given points A, B, C.
35. To find the centre of a circle from a given arc AC.
36. At the given equidistant poihtS A, B, C, D; etc.. on a given arc,
to draw a numher of radial lines, the centre of the circle being
inaccessible.
With the points A, B, C,
D, etc., as centres, with radii
larger than a division, des
cribe arcs cutting each other
at E, F, etc. Thus, from A
and C as centres, describe
arcs cutting each other at E,
Fig 61. J T^ .u r
and so on. Draw the Imes
BE, CF, etc., which will be the radial lines required.
37. To draw the arc of a
circle through three
given points A, B, C,
the centre of the circle
being inaccessible.
With A and C as
centres, and with a radius
equal to AC, draw inde
finite arcs. From the
points C and A draw lines
through B till they meet
Fig. 62.
PROBLEMS. ?.;
the arcs in D and E. From D and E set off short equal
distances on the arcs above and below ; join the divisions
on arc AD to point C, and the divisions on arc CE to
point A. Where the lines from corresponding points intersect,
we obtain a point in the arc ; for instance, where the line from
Ei intersects the line from Di, we get the point F ; and so on
with the other points, by joining which with a fair curve, we get
the required arc.
EXERCISES.
' 1. Draw two parallel lines 2" long and if" apart.
• 2. Draw a line 3.75" long; at the righthand end erect a perpen
dicular 2.25" high ; then, 1.50" from it, another perpendicular i" high ;
and bisect the remaining length by a line 3" long, at right angles to it.
^ 3. Draw a line 3" long, and divide it into seven equal parts.
4. Draw a line 2f " long ; from the lefthand end mark off a distance
■equal to i^", and from the righthand end a distance of g" ; draw
another line 1.75" long, and divide it in the same proportion.
5. Mark the position of three points A, B, and C — A to be if"
from B, B to be 2^" from C, and C i" from A ; and join them.
6. Draw an angle equal to the angle ACB in the preceding question,
and bisect it.
7. Draw a right angle, and trisect it ; on the same figure construct
and mark the following angles, viz. 15°, 30°, 45°, 60", 75°; and also
74°, 22\\ and y]\\
8. Construct a triangle with a base l" long ; one angle at the base
to be 60°, and the side opposite this angle 2" long.
9. Construct a triangle with a base 2.25" long, and altitude o't
I75"
10. On a base i" long, construct an isosceles triangle ; the angle at
its vertex to be 30°.
11. Draw a scalene triangle on a base 2" long ; and construct a
similar triangle on a base 1.75" long.
12. On a base 2^" long draw a triangle with the angle at its vertex
90°.
13. Let a line 2.25" long represent the diagonal of a rectangle ;
complete the figure, making its shorter sides " long.
14. Construct a rhombus with sides if" long, and one of its angles
60°.
15. Draw any two converging lines, and through any point between
them draw another line which, if produced, would meet in the same
point as the other two lines produced.
28
GEOMETRICAL DRAWING AND DESIGN.
16. Fix the position of any three points not in the same line, and
draw an arc of a circle through them.
17. Draw an arc of a circle, and on it mark the position of any three
points ; from these points, without using the centre, draw lines which,
if produced, would meet in the centre of the circle containing the arc.
18. Construct a triangle, the perimeter to be 5.6", and its sides in
the proportion of 5 : 4 : 3.
19. Draw two lines AB, AC containing an angle of 75''. Find a
point P, f" from A 13 and ^" from AC. Complete the isosceles triangle,
of which BAC is the vertical angle, and the base passes through P.
(June, '00.)
20. Draw the figure shown (Fig. 63) according to the given figured
dimensions. (May, '97.)
m
\VV7
Fig. 64.
21. Draw the pattern (Fig. 64) according to the given dimensions.
(April, '96.)
22. Copy the diagram (Fig. 65) enlarging it to the dimensions
figured. (April, '98.)
23. Draw the figure shown in the diagram, (Fig. 66) making the side
of the outer hexagon i^" long. (June, '98.)
''^Qll
E
m
Fig. 66.
Fig. 67.
Fig. 68.
24. Draw the corner ornament shown (Fig. 67), using the figured
dimensions, (April, '99.)
25. Draw the given diagram of window tracery (Fig. 68) using the
figured dimensions. The arch is ' equilateral'. (June, '99.)
PROBLEMS.
29
26. Draw the given outline of window tracery (Fig. 69) using the
figured dimensions. The arch is " equilateral," and all the arcs are of
equal radius. (April, '00.)
/XX ;
11
n^r
C
A
A
A
"^
f*.
>fs
1 1
_J L_
nX,'
Fig. 69.
Fig. 70.
27. Draw the given frame (Fig. 70), using the figured dimensions.
The border is J" wide throughout. (April, '00.)
28. Draw the given figure (Fig. 71), using the figured dimensions.
(This problem is intended as an exercise in the use of T and set squares.)
(June, '00.)
 z'  ■*■
Fig. 71.
• '  J*' .»
Fig. 72.
29. Draw the given figure (Fig. 72), using the figured dimensions.
The spaces are ^" wide throughout. (June '99. )
30. Construct an isosceles rightangled triangle having its hypotenuse
(or side opposite the rightangle) 2" long, ^^'ithin it inscribe a square
having one of its sides in the hypotenuse of the triangle. Measure and
state, as accurately as you can, the length of one side of the square.
(June, '01.)
C.
Q. 32.
31. The given figure (Q. 31) is made up of a rectangle and semi
circles. Make a copy of it, using the figured dimensions. (June, '02.)
32. Make a figure similar to the given figure (Q. 32) and having the
height CD increased to 2f ". The centres of the arcs are given.
(June, '03.)
CHAPTER III.
POLYGONS.
Regular Polygons are figures that have equal sides and equal
angles. To construct a regular polygon, we must have the length
of one side and the number of
sides ; if it is to be inscribed in
a circle, the number of its sides
will determine their length.
If we take any polygon, regu
lar or irregular, and produce all
its sides in one direction only,
Fig. 73, we shall find that the
total of all the exterior angles,
shown by the dotted curves, is
equal to 360°, or four right
angles ; and if we join each
angle of the polygon to any
point in its centre, the sum of
the angles at this point will also
be 360°, and there will be as
many angles formed in the centre
as there are exterior angles.
In regular polygons these angles at the centre will, of course,
be equal to each other ; and if we produce the sides in one
direction, as in Fig. 73, the exterior angles will be equal to each
other; and as the number of angles at the centre is equal to the
number of exterior angles, and the sum of the angles in each
instance is equal, the angle at the centre must equal the exterior
angle.
POLYGONS.
iiven line AB, for
To construct any regular polygon on the
example, a nonagon,
360° r 9 =40°.
So if we draw a line at B,
making an angle of 40° with
AB produced, it will give us
the exterior angle of the nona r
gon, from which it will be
easy to complete the polygon.
The perimeter of a polygon is sometimes given, e.g. Construct
an octagon the perimeter of which is 6 inches.
^.i
B
• 74
inches.
inches, inches.
 75 = I
Draw the line AB this length. It has been shown that the
exterior angles and those at the centre of a regular polygon
are equal, 360° ^ 8 = 45°. Pro
duce the Hne AB, and con
struct an angle of 45°. Make
BC = AB. We now have three
points from which we can
draw the circle containing the
required polygon (Prob. 33).
The polygon could also be
drawn, after finding the length of AB, by any of the methods
shown for constructing a polygon on a given straight line.
The centre of any regular polygon is the centre of the circle
that circumscribes it.
Any regular polygon can be inscribed in a circle by
making angles at the centre equal to the exterior angle as
above.
If tangents to a circle circumscribing a regular polygon be
drawn parallel to the sides of the inscribed polygon, or if
tangents be drawn at the angles of the inscribed polygon, a
similar figure will be described about the circle, and the
circle will also be " described by," i.e. contained in, a similar
figure.
In the following problems two general methods are given for
constructing regular polygons on a given line, and two for
32
GEOMETRICAL DRAWING AND DESIGN.
inscribing them in a given circle ; but as these general methods
require either a line or an arc to be first divided into equal
divisions, the special methods for individual polygons are pre
ferable, which are also given.
See also how to construct any angle without a protractor
(Prob. 134), and its application to polygons.
38. To inscribe in a circle, a triangle, square, pentagon, hexagon,
octagon, decagon, or duodecagon.
Describe a circle, and draw the two diameters AEand BD at
right angles to each other ; join BA. AB is a side of a square.
Set off on circumference, AF
equal to AC ; AF is a side of a
hexagon ; join AF and EF ; EF
is a side of an equilateral
triangle. With D as centre,
and radius equal to EF, mark
off G on EA produced. With
G as centre, and radius equal to
AC, set off H on circumference;
join AH ; AH is a side of an
octagon. With D and E as
centres, and radius equal to
DC, set off the points I and J on
circumference ; join ID ; ID is
a side of a duodecagon. With
CG as radius, and J as centre, mark off K on diameter BD.
With CK as radius, mark off on circumference from B the points
L and M ; join BL and BM ; then BL is the side of a decagon,
and BM is a side of a. pentagon.
Approximate Constructions, 39 42.
39. To inscribe any regular polygon in a given circle;
for example, a heptagon.
Describe a circle, and draw the diameter AB. Divide AB
POLYGONS.
33
into as many equal divisions as there are sides to the polygon
(in this instance seven). With
A and B as centres, and with
radius equal to AB, describe
arcs intersecting at C. From C
draw the line CD, passing through *
the second division from A,^ till
it meets the circumference at D.
Join AD, which will give one side
of the polygon ; to complete it,
mark off AD round the circum
ference.
This method of constructing
polygons is due to the Chevalier
Antoine de Ville (1628), and
although useful for practical purposes, is not mathematically
correct.
Fig. 77.
40. To inscribe any regular polygon in a given circle
(second method) ; for example, a nonagon.
Describe a circle, and draw the radius CA. At A draw a
tangent to the circle. With
A as centre, and with any
radius, draw a semicircle, and
divide it into as many equal
parts as there are sides to
the polygon (in this instance
nine).
From point A draw lines
through each of these divisions
till they meet the circum
ference. Join these points,
which will give the polygon
required.
1 Whatever number of sides the polygon may have, the Hne CD is ahvays
drawn through the second division from A.
C
34
GEOMETRICAL DRAWING AND DESIGN.
41. On a given line AB, to describe a regular polygon;
for example, a heptagon.
Produce AB to C. With A as centre, and AB as radius,
describe a semicircle, and
divide it into as many equal
divisions as there are sides to
the polygon (in this instance
seven). Join A with the second
division from C in the semi
circle,^ which will give point
D. Join AD. Through the
points DAB describe a circle.
Set off the distance AD
round the circumference, and
join the points marked, which will give the required polygon.
42. On a given line AB, to describe a regular polygon
(second method) ; for example, a pentagon.
At point B erect a perpendicular equal to AB. With B as
centre, and radius BA, draw the
quadrant AC, and divide it into
as many equal divisions as
there are sides to the polygon
(in this case five).
Join the point B with the
second division from C.^ Bisect
/ AB at D, and erect a perpendi
/ cular till it meets the hne drawn
from B to the second division,
which will give point E. From
E as centre, and with radius
EB, describe a circle. From
AB, mark off round the circum
B
Fig. 8o.
point A, with distance
1 The line AD is always drawn to the second division from C, whatever
number of sides the polygon may contain.
POLYGONS.
35
ference the points of the polygon,
will give the pentagon required.
Join these points, which
43. On a given straight line AB, to construct a regular
pentagon. (True construction.)
At B erect BC perpen
dicular to AB, and equal
to it. Bisect AB in D.
With D as centre, and DC
as radius, draw the arc CE
meeting AB produced in E.
With A and B as centres,
and radius equal to AE,
draw arcs intersecting at
F. With A, B, and F as
centres, and radius equal
to AB, draw arcs intersect
ing at H and K. Join AH,
HF, FK, and KB. Then
required.
AHFKB will be the pentagon
44. On a given straight line AB, to construct a regular hexagon.
With A and B as centres,
and radius AB, draw the arcs
intersecting each other at C.
With C as centre, and with
the same radius, draw a circle.
With the same radius com
mencing at A, set off round
the circle the points D, E, F,
G. Join AD, DE, EF, FG,
and GB, which will give the
hexagon required.
E.'. f
Fig. 82.
36
GEOMETRICAL DRAWING AND DESIGN.
45. In a given circle to inscribe a regular heptagon
(approximately).
Draw any radius AB, and
bisect it in C. Through C
draw DE perpendicular to
AB. With CD as radius,
commencing at E, set off
round the circle the points
F, G, H, K, L, and M, by
joining which we get the
required heptagon.
Fig. 83.
46. On a given line AB, to construct a regular heptagon
(approximately).
With B as centre, and BA as radius, draw a semicircle
meeting AB produced in
C. With centre A, and
radius AB, draw an arc
cutting the semicircle in
D. Draw DE perpendi
cular to AB. With C as
centre, and radius equal
to DE, draw an arc cut
ting the semicircle in F.
Join BF. From the three
points A, B, F, find the
centre of the circle H
(Prob. 33). With H as
centre, and radius HA,
Fig. 84.
draw a circle. With AB as radius commencing at F, set off
on the circle the points K, L, M, and N, by joining which we
get the heptagon required.
POLYGONS.
37
47. On a given line AB, to construct a regular octagon.
At A and B erect the perpendiculars AC and BD. Produce
AB to E. Bisect the
angle DBE by the line
BF. Make BF equal to
AB. From F draw the
line FH parallel to AB,
and make KH equal to
LF. Join AH. Set off
on AC and BD, from
points A and B, a length
equal to H F, which will
give the points C and D.
With the points C, D, H,
and F as centres, and a
radius equal to AB, draw ^, g.
arcs intersecting at M
and N. Join HM, MC, CD, DN, and NF, which will give the
octagon required.
48. In a given circle to inscribe a nonagon.
Draw the diameters AB and CD perpendicular to each other.
With A as centre,
and radius AE, draw
the arc cutting the
circle in F. With B
as centre, and radius
BF, draw the arc
cutting CD produced
in G. With G as
centre, and radius
GA, draw the arc
cutting CD in H.
With HC as radius,
commencing at A,
set off on the circle
the points K, L, M,
N, O, P, Q, and R, by joining which we get the nonagon required.
C
3^
GEOMETRICAL DRAWING AND DESIGN.
49. In a given circle to inscribe a regular undecagon.
Draw the two diameters AB and CD cutting each other in E.
With A as centre, and AE
as radius, draw an arc cut
ting the circle in H. With
D as centre, and DE as
radius, draw an arc cutting
the circle in F. With F as
centre, and FH as radius,
draw the arc cutting CD in
K. Join HK. Then HK
is equal to one side of the
undecagon. With HK as
radius, starting from H, set
off en the circle the points of
the required undecagon, and join them.
EXERCISES.
1. Inscribed in circles of varying diameters, draw the regular
polygons, from a pentagon to a duodecagon, by a general method and
figure the angles formed by their sides.
2. In circles of various radii, draw all the preceding polygons by
special methods. Join their angles to the centre of the circle by radii,
and figure the angles between the radii.
3 On lines varying in length, draw the same polygons by a general
method.
4. Construct on lines of different lengths the same polygons by
special methods.
5. Construct an irregular hexagon from the following data : Sides,
AB I", BC if, CD f", DE ih", EF i" ; Angles, ABC 140°, BCD
130°, CDE 110°, DEF 120°.
6. Construct an irregular pentagon from the following data : Sides,
AB 1.25", BC 1.3", CD 1.2", DE 1.3", EA 1.4"; Diagonals, AC
1.8" AD 1.6".
7. Construct a regular polygon with one side i" in length and one
angle 140°.
8. How many degrees are there in each of the angles at the centre
of a nonagon ?
9. Construct a regular polygon on the chord of an arc of 72°.
POLYGONS.
39
10. Inscribe in any given circle an irregular heptagon whose angles
at the centre are respectively 52°, "j^^^ 45°. 63°, 22°, 36'', and 69°.
11. Construct a regu]ar octagon of i" side, and a second octagon
having its angles at the middle points of the sides
of the first. (May, '97.)
12. Draw a figure similar to the one shown
(Fig. 88), the points of the star being at the angles
of a regular heptagon inscribed within a circle of
1 1" radius. (June, '97.)
'3. Within a circle of i^" radius inscribe a
regular nonagon. Within tne nonagon inscribe a
rectangle having all its angles in the sides of the
nonagon, and one of its sides i^" long.
14. Construct a regular pentagon of iV side. Describe five circles
off" radius, having their centres at the five angles of the pentagon.
(June, '98.)
15. Draw the given figure (Q. 15), making the radius of the outer
circle i\". (April, '01.)
16. Construct a regular pentagon of 2" side, and a similar pentagon
of 2" diagonal. The two figures should have the same centre.
(N.B. — The protractor may not be used for obtaining the angle of
the pentagon.) (June, '02.)
Fig. £8.
(April, 98.)
Q. 15
Q 19.
17. Copy the diagram (Q. 17), using the figured dianensions.
(June, '02.)
18. Within a circle of I "5" radius inscribe a regular nonagon. With
the same centre describe a regular nonagon of 075" side. (May, '03.)
19. Copy the given figure (Q. 19). The straight lines are to form a
regular pentagon of I 'S" side, and the five equal segments are to have
their arcs tangential at the angles of the pentagon. Show all your
constructions clearly. ' (June, '03.)
CHAPTER IV.
INSCRIBED AND DESCRIBED FIGURES.
50, To inscribe an equilateral triangle in a given circle ABC.
Find the centre E (Prob.
H
32), and draw the diameter DC.
With D as centre, and DE as
radius, mark off the points A
and B on the circumference of
the circle. Join AB, BC, and
CA. Then ABC will be the
inscribed equilateral triangle
required.
51. To describe an equilateral
triangle about a given circle
ABC.
At the points A, B, and C
draw tangents to the circle
(Prob. 84), and produce them till they meet in the points F, G, and
(^ H. Then the equilateral triangle
FGH will be described about
the circle ABC, as required.
52.
To inscribe a circle in a
given triangle ABC.
Bisect the angles CAB and
ABC by lines meeting in D.
From D let fall the line DE,
perpendicular to AB. With D
as centre, and DE as radius,
inscribe the circle required.
INSCRIBED AND DESCRIBED FIGURES.
41
53. To describe a circle about
a given triangle ABC.
Bisect the two sides AB
and AC perpendicularly by
lines meeting in D. With D
as centre, and DA as radius,
describe the required circle.
54. To describe an equilateral
triangle about a given
square ABDC.
With C and D as centres,
and with CA as radius, de
scribe arcs cutting each other
at E. With E as centre, and
with the same radius, mark
off the points F and G on
these arcs. Join CF and DG,
and produce them till they
meet in the point H, and the
base AB produced in K and
L. Then HKL is the re
quired equilateral triangle.
55. In a given triangle ABC,
to inscribe an oblong hav
ing one of its sides equal
to the given line D.
From A, along the base
AB, set off AE equal to the
given line D. From E draw
the line EF parallel to AC.
Through F draw the line FG
parallel to the base AB.
From G and F draw the lines
GH and FL perpendicular to
AB. GFLH is the oblong
required.
Fig. 93
42
GEOMETRICAL DRAWING AND DESIGN.
To inscribe a square in a given circle.
C y\ Draw any two diameters AB
and CD at right angles to each
other. Join the extremities of
these diameters. ACBD is the
inscribed square required.
57. To describe a square about
a given circle.
At the points A, C, B, D draw
tangents meeting each other
at the points E, F, G, H. (Prob.
84). EFHG is the described
square required.
To inscribe a circle in a given square.
Draw the diagonals AB and
CD intersecting each other at
E. From E draw EF perpendi
cular to AD. With E as centre,
and EF as radius, draw a
circle. This will be the in
scribed circle required.
59. To describe a circle about a
given square.
With centre E, and radius
EA, draw a circle. This will be
the described circle required.
60. To inscribe a square in
a given rhombus.
Draw the two diagonals
AB and CD. Bisect the
angles AEC, AED, and
^ B produce the lines each way
till they meet the sides of
the rhombus in the points
F, G, H, L. Join FG, GL,
LH,andHF. FGLH is the
inscribed square required.
INSCRIBED AND DESCRIBED FIGURES.
43
61. To inscribe a circle in a
given rhombus.
Draw the diagonals AB
and CD intersecting each
other in E. From E draw
the line EF peipendicular to^
AD (Prob. 7). With E as
centre, and EF as radius,
draw a circle. This is the
inscribed circle required.
62. To inscribe an equilateral
triangle in a given square
ABDC.
With B as centre, and radius
BA, draw the quadrant AD ;
and with same radius, with A
and D as centres, set off on
the quadrant the points F and
E. Bisect AE and FD, and
through the points of bisection ■^~
draw lines GB and HB, cut
ting the given square in G and
H. Join GH. BGH is the
inscribed equilateral triangle
required.
63. To inscribe an isosceles
triangle in a given square
ABDC, having a base
equal to the given line E.
Draw the diagonal AD.
From A, along AD, set off
AF equal to half the given
base E. Through F draw
the line GH perpendicular to
AD (Prob 5). Join GD and
HD. GDH is the inscribed
isosceles triangle required.
Fig. 99.
44
GEOMETRICAL DRAWING AND DESIGN.
64. To inscribe a square in a given trapezium ACBD which
has its adjacent pairs of sides equal.
Draw the two diagonals
AB and CD. From point C
set off CE perpendicular and
equal to CD. Join EA by a
line cutting CB in G. Draw
the hne GF parallel to AB.
From points F and G draw
the lines FH and GK parallel
to CD. JoinHK. FGKH is
the inscribed square required.
65. To inscribe a circle in a
given trapezium ACBD
which has its adjacent
pairs of sides equal.
Bisect any two adjacent
angles, as ADB and DBC,
by lines meeting in E. From
E draw EF perpendicular to
CB. With E as centre, and
EF as radius, draw a circle.
This will be the inscribed
circle required.
To insert a rhombus in a given rhomboid ABDC.
>^ Draw the diagonals AD
C ,'P P and CB intersecting at E.
Bisect two adjacent angles,
as CED and DEB, by lines
cutting the given rhomboid
in F and G, and produce
these lines to K and H.
Join FG, GK, KH, and
HF. FGKH is the in
scribed rhombus required.
67. To inscribe an octagon in a given square ABDC.
Draw the two diagonals AD and CB intersecting each other
in E. With A as centre, and AE as radius, mark off the points F
INSCRIBED AND DESCRIBED FIGURES.
45
and G on the sides ot the
square. Proceed in the same
manner with the angles B, C,
and D as centres, which will
give the eight points required
on the given square, by joining
which we obtain the required
inscribed octagon.
68. To inscribe a square in a
given hexagon ABCDEP.
Draw the diagonal EB ;
bisect it in G, and draw HK
perpendicular to it. Bisect
two adjacent angles, as BGK
and EGK, by lines meeting
the hexagon in O and M.
Produce these two lines till
they meet the opposite sides
of the hexagon in L and N.
Join LM, MO, ON, and NL.
LM ON is the inscribed square
required.
69. To inscribe four equal
circles in a given square
ABDC ; each circle to touch
two others, as well as two
sides of the given square.
Drav.' the two diagonals
AD and CB intersecting at E.
Bisect the sides of the square
in the points F, G, H, and L
(Prob.i). Draw FH and GL.
Join FG, GH, HL, and LF,
which will give the points M,
O, P, and R. From M draw
MN parallel to AB. With
M, O, P, and R as centres, and
radius equal to MN, describe
the four inscribed circles re
quired.
46
GEOMETRICAL DRAWING AND DESIGN.
70. To inscribe four equal circles in a given square ABDC ;
each circle to touch two others, and one side only of the
given square.
Draw the two diagonals
AD and CB intersecting in
the point E. Bisect the sides
of the square in the points
F, G, H, and L. Join FH
and GL. Bisect the angle
GDE by a hne meeting GL
in M. With centre E, and
radius EM, mark off the
points N, O, and P. With
the points M, N, O, and P as
centres, and radius equal to
MG, describe the four in
scribed circles required.
71. To inscribe three equal circles in a given equilateral triangle
ABC ; each circle to touch the other two, as well as two
sides of the given triangle.
^ Bisect the two sides of the
triangle at right angles by
lines meeting at the centre D.
Draw a line from point C
through the centre D till it
meets the base at E. Bisect
the angle BEC by a line
meeting DB in F. With D
as centre, and DF as radius,
mark off the points K and
L. From F draw the line
FH perpendicular to AB.
With the points F, K, and L
as centres, and with a radius
equal to FH, draw the three inscribed circles, required.
72. In a given equilateral triangle ABC, to inscribe three equal
circles touching each other and one side of the triangle only.
Bisect two sides of the triangle at right angles by lines
INSCRIBED AND DESCRIBED FIGURES.
47
meeting at the point D. Join CD and produce it to E,
Bisect the angle DBE by a line cutting CE in F. With D as
centre, and DF as radius, mark off the points G and H,
From the points F, G, and
H as centres, and with a .^
radius equal to FE draw
the three inscribed circles
required.
73. In a given equilateral
triangle ABC, to inscribe
six equal circles touching
each other.
Having drawn the three
circles according to the last
problem, draw lines through
G and H, parallel to the sides of the triangle, till they meet the
lines bisecting the angles in the points L, M, and N. These
points are the centres of the three circles which will complete
the six inscribed circles required.
74. In a given octagon to inscribe four equal circles touching
each other.
Draw the four diagonals
meeting in C. Bisect the
angle ABC by a line inter
secting AC in D. With C
as centre, and CD as radius,
mark off the points F, G, and
H. From D draw the line
DL perpendicular to KC.
With D, F, G, and H as
centres, and a radius equal
to DL, draw the four in
scribed circles required.
Fig. log
48
GEOMETRICAL DRAWING AND DESIGN.
75. In a given circle to draw four equal circles touching
each other.
Find centre of circle E
(Prob. 32). Draw the two
diameters AB and CD at
right angles to each other.
At A and D draw tangents
to the circle to meet at the
) point F (Prob. 84). Join FE.
Bisect the angle EFD by a
line cutting CD in G. With
E as centre, and EG as
radius, mark off the points
H, K, and L. With G, H,
K, and L as centres, and
with a radius equal to GD,
draw the four inscribed circles required.
76. In a given circle to inscribe any number of equal circles
touching each other. For example, five.
Find the centre C (Prob. 32). Divide the circumference into
five equal parts (Prob. 38), and draw the five radii to meet the
circumference in the points
M, N, O, P, and R. Bisect
the angle MCN by a line
meeting the circumference in
A. Through A draw a line
at right angles to CE till it
meets the Hnes CM and CN
produced in B and D. Bisect
the angle DBC by a Hne to
meet CA in E. With C as
centre, and CE as radius,
draw a circle, and bisect
each arc on this circle,
between the five radii, in
the points F, G, H, and L.
With E, F, G, H, and L as centres, and a radius equal to EA,
draw the five inscribed circles required.
INSCRIBED AND DESCRIBED FIGURES.
49
77. About a given circle A, to describe six circles equal to it,
touching each other as well as the given circle.
Find centre of circle A (Prob.
32). With A as centre, and a
radius equal to the diameter of
the given circle, draw the circle
BCDEFG. Draw the diameter
BE. Take the diameter of the
giv^en circle and mark off from
E the points D, C, B, G, and
F. With each of these points
as a centre, and a radius equal
to that of the given circle, draw
the six circles required.
Fig. 112.
EXERCISES.
1. Within a given circle inscribe a square, and about the same
circle describe an equilateral triangle.
2. Construct a rhombus with sides i^" long, and its shorter
diagonal 1.75" ; inscribe a circle within it, and let the circle circum
scribe an equilateral triangle.
3. Construct a trapezium with two of its sides if" and two i"
respectively, and with its longer diagonal 2^" ; inscribe within it a
square, and let the square circumscribe an equilateral triangle.
4. Draw a regular hexagon with i" sides and let it circumscribe a
square ; inscribe a regular octagon within the square.
5. Draw any triangle, and describe a circle about it.
6. Construct a square of 2^" sides, and in it inscribe an isosceles
triangle with a if" base ; inscribe within the triangle a rectangle, one
side of which is i".
7. Within a square of 1.75" sides, inscribe an isosceles triangle with
angle at vertex 60° ; inscribe a circle within the triangle.
8. Within a circle of any radius, inscribe a regular duodecagon, and
let it circumscribe a hexagon.
9. Construct a rhomboid with sides 2" and i^", its contained angle
to be 60° ; inscribe within it a rhombus.
10. Within an equilateral triangle of 3" sides, inscribe a circle, and
within it 3 equal circles.
50 GEOMETRICAL DRAWING AND DESIGN.
11. Draw a circle of i" radius ; inscribe within it an equilateral
triangle ; inscribe within the triangle three equal circles touching each
other and each one side of the triangle only.
12. Construct a square of 2" sides, and let it circumscribe four equal
circles ; each circle to touch two others, as well as two sides of the
square.
13. Construct a square with sides of 2.3", and inscribe four equal
circles within it ; each circle to touch two others, as well as one side
only of the square.
14. Within a triangle of 2.7" sides, inscribe six equal circles.
15. • Within a circle of 1.7" radius, inscribe seven equal circles.
16. Draw two concentric circles, and between them, six equal circles,
to touch each other as well as the two concentric circles.
17. In a decagon, inscribe five equal circles.
18. Construct an equilateral triangle of 2f" side. Bisect all three of
its sides and join the points of bisection. Within each of the four
equilateral triangles thus formed inscribe a circle. (April, '98.)
19. Draw an equilateral triangle, a scalene triangle, a rightangled
triangle, and an oblong ; a trapezium, and a regular polygon of
eleven sides, each in a 2inch circle, and write the names to each.
(May, '96.)
20. Draw each of the following figures in a separate 2inch circle, an
isosceles triangle, an obtuseangled triangle, an acuteangled triangle, a
square, a rhombus, and a rhomboid, and write the name to each.
(May, '97.)
21. Draw a square^ an ob.'ong, and a trapezium ; a pentagon^ a
hexagon, an octagon, and t7U0 parallel straight lines, each in a
separate 2inch circle, and write the name to each. (April, '98.)
22. About a square of \" side describe a triangle having one of its
angles 60° and another 70°. (May, '97.)
23. Within a rhombus, sides 2", one angle 60°, inscribe an ellipse
touching the sides of the rhombus at iheir middle points. (June, '97.)
24. Construct a rectangle 2V' x if". Within it inscribe two other
rectangles, each similar to the first, concentric with it and having their
longer sides ij" and i" long respectively. (April, '98.)
25. Within an equilateral triangle of 3" side inscribe three equal
circles each touching the two others, and tivo sides of the triangle.
(May, '97.)
26. Construct a square of i" side, and within it inscribe a rectangle
having one of its angles on each side of the square and one of its sides
i"long. (June, 97.)
27. Construct a triangle, sides i^", 2", and 2^", and within it
INSCRIBED AND DESCRIBED FIGURES. 51
inscribe an equilateral triangle having its three angles in the three
sides of the first triangle. (June, '98.)
28. Within a square of if" side inscribe a regular octagon having all
its angles in the sides of the square, (April, '99.)
29. Construct a regular heptagon of i" side, and within it inscribe an
equilateral triangle. (June, '99.)
30. Construct a quadrilateral ABCD from the following data : —
Sides— AB=ii", BC=i4"
Angles— ABC =105°, BAD = 75°
The four angles of the figure all lie in the circumference of a circle.
(April, '98.)
31. Within a circle of i" radius inscribe a regular pentagon.
About the same circle describe another regular pentagon, having its
sides parallel to those of the inscribed pentagon. (April, '96.)
32. Within a circle of 2" diameter, inscribe four equal circles each
touching the given circle and two of the others. (June, '98.)
33. Within a circle of i" radius inscribe a regular heptagon. Draw
a second similar heptagon, of which the longest diagonals are 2" long.
(April, '99.)
34. Within a circle of i" radius, inscribe a regular hexagon.
Within the hexagon inscribe three equal circles touching each other,
and each touching two sides of the hexagon. (June, '99.)
35. Construct a rhombus having its sides 2" long and one of its
angles 75°. Within it inscribe two equal circles touching each other,
and each touching two sides of the rhombus. (April, '99. )
36. Within a circle of 2" diam. inscribe a regular pentagon. Draw
also a second regular pentagon concentric with the first one, its sides
being parallel to those of the first and I5" long. (June, 1900.)
37. The sides of a triangle are i", i^" and i^" long. About this
triangle describe a circle, and about the circle describe a triangle of the
same shape as the given one. The points of
contact must be found and clearly shown.
(April, 1900.)
38. Describe a circle touching both the
given circles (Fig. 113) and passing through
the point P on the circumference of the
larger one, three times the size of figure.
(April, 1900.)
39. About a circle of i" diam. describe Fig. 113.
six equal circles, each touching the given
circle and two of the others. Then describe a circle touching and
enclosing all six of the outer ones.
52
GEOMETRICAL DRAWING AND DESIGN.
40. Make an enlarged copy of the given diagram (Fig. 114), using
the figured dimensions. (June, '98. )
Fig. 115.
41. Copy the diagram (Fig. 115) according to the given dimensions.
Show clearly how the centre of the small circle is determined.
(June, '98.)
42. Within a regular hexagon of ii"side, inscribe a square having
all its angles in the sides of the hexagon. Within the square inscribe
four equal circles, each touching two of the others and two sides of the
square. (April, '01.)
43. About a circle of 075" radius describe an equilateral triangle.
Describe three equal circles touching the given circle and having their,
centres at the angles of the triangle. Determine the points of contact.
(April, '02.)
44. Draw the given figure (Q. 44), making the side of the square 2^"
long. Show clearly how all the points of contact are determined.
(April, '02.)
Q. 44.
46. About a circle or o*8" radius describe a rhombus having one oi
its angles 54°. Each side of the rhombus touches the circle. Deter
mine the four points of contact. (June, '02. )
46. Describe a semicircle of I '5" radius. Within it inscribe two
circles, each touching the other, and also touching the circumference
and diameter of the semicircle. (June, '03.)
CHAPTER V.
FOILED FIGURES.
Foiled figures are constructed on the regular polygons, and
are of two kinds : viz. those having tangential arcs, and those
having adjacent diameters.
Problem 78 is an illustration of the former kind. The arcs
simply touch, and if produced would not intersect each other.
The angles of the polygons are the centres of the circles con
taining the arcs.
Problem 79 is an illustration of the latter kind. The sides of
th polygon form the diameters of the semicircles, the centre
each side being the centre of the circle containing the
arc. If these slyc^ were produced, they would intersect each
other.
78. To construct a foiled figure
about any regular polygon,
having tangential arcs. For
example, a hexagon.
The Hexafoil. — Bisect one side
AB in C. With each of the
angular points as centres, and
AC as radius, draw the six arcs,
as required.
Fig. 116.
54
GEOMETRICAL DRAWING AND DESIGN.
79. To construct a foiled figure about any regular polygon, having
adjacent diameters. For example, a pentagon.
Tlie Cinquefoil. — Bisect
each side of the pentagon
in the points C, D, E, F, and
G. With each of these points
as centres, and with a radius
equal to CA, draw the five
arcs required.
Note. — If these arcs are
to have a stated radius, the
length of the hne AB, in each
instance, will be twice the
required radius.
In a given equilateral triangle ABC, to inscribe a trefoil.
The Trefoil. — Bisect the
angles CAB and ABC by
lines produced till they meet
in L, and the sides of the
triangle in D and E. From
C, through centre L, draw the
line CF. Join DE, EF, and
FD. From G draw GI per
pendicular to AC. WithG,H,
and K as centres, and a radius
equal to GI, draw the three
arcs till they meet each other,
which will form the trefoil
required.
81. Within a given circle, to inscribe three equal semicircles having
adjacent diameters.
Find centre of circle A (Prob. 32). Draw the diameter BC, and
FOILED FIGURES.
55
the radius AD perpendicular to it (Prob. 5). Trisect the angle
BAD in E and F (Prob. 13).
Set off DG equal to DF.
Join FA and GA by lines
produced to L and H. Join
EG by a line cutting FL in
M. With A as centre, and
AM as radius, set off the
points N and O. Join MN,
NO, and OM, which are the
diameters of the required
semicircles. With R, P, and
S as centres, and a radius
equal to RM, draw the three
semicircles required.
82. In a given square ABDC, to inscribe four semicircles having
adjacent diameters.
The Quatrefoil. — Draw the
diagonals AD and CB.
Bisect each side of the square
in the points E, F, G, and
H, and join EF and GH.
Bisect HD in K and FB in
L, and join KL, cutting GH
in N. With P as centre, and
PN as radius, mark off the
points M, O, and R, and
join NM, MO, OR, and RN,
which will form the diameters
of the required semicircles.
With S, T, U, and V as
centres, and with a radius
equal to SN, draw the four
semicircles required.
56
GEOMETRICAL DRAWING AND DESIGN.
83. Within a given circle to inscribe any number of equal semi
circles having adjacent diameters. For example, seven.
The Heptafoil. — Find centre C (Prob. 32). Divide the circle
into as many equal parts
as the number of semi
circles required (in this
case seven), and from
each of these points, A,
B, D, E, F, G, and H,
draw diameters. From
the point A draw the
tangent AM (Prob. 84),
and bisect the angle CAM
by a Hne cutting CK in
N (Prob. 12). With C as
centre, and CN as radius,
mark off the points O, P,
R, S, T, and U. Join each
of these points to form the polygon. From the centre of each
side of the polygon, with a radius equal to half of one of its
sides, draw the seven semicircles required.
EXERCISES.
1. Construct an equilateral triangle of i^" sides, and about it describe
a trefoil having tangential arcs.
2. Construct a pentagon of f" sides, and about it describe a cinque
foil having adjacent diameters.
3. Draw a pentagon of ^" sides, and about it construct a cinquefoil
having tangential arcs.
4. In a circle of i^" diameter, draw nine equal semicircles having
adjacent diameters.
5. In a circle of 1" diameter, inscribe a quatrefoil having tangential
arcs.
6. Construct a regular decagon in a circle of " radius, and within it
inscribe a cinquefoil.
Note. — Foiled figures can be inscribed in all the regular polygons
that have an even number of sides, by first dividing them into trapezia
and then proceeding by the method shown in Prob. 65 ; or they can be
drawn in circles divided into any number of equal sectors, by Prob. 106.
CHAPTER VI.
TANGENTS AND TANGENTIAL ARCS
84. To draw a tangent to a
given circle at a given
point A.
Find the centre of the"
circle C (Prob. 32). Join AC.
From A draw the hne AB
perpendicular to AC, and
produce it. Then AB is the
tangent required.
Fig. 122
85. To draw a tangent to a
given circle from a given
point A outside it.
Find the centre C (Prob.
32). Join AC, and bisect
in B. From B as centre,
and with BA as radius, de
scribe" a semicircle cutting
the circumference in D. Join
AD, and produce it. Then
the line AD is the tangent
required.
Fig. 123.
58
GEOMETRICAL DRAWING AND DESIGN.
86. To draw a tangent to the
arc of a circle at a given
point A without using the
centre.
Draw the chord AB and
bisect it in C). At C erect a
perpendicular to AB cutting
the arc in D. Join AD.
Make the angle DAE equal
to DAC. Then EA produced
is the tangent required.
87. To draw a circle with
radius equal to line D to
touch two straight lines
forming a given angle
ABC.
Bisect the angle ABC by
the line BE. Draw the hne
FIT parallel to BA (Prob. 4),
and at a distance equal to
given line D from it. Where
FH intersects BE will be the
centre of the circle. With
E as centre, and D as radius,
draw the circle required.
88. To draw tangents to a
circle from a given point
A, the centre of the circle
not being known.
From point A draw any
three secants to the circle.
as CB, DE, and GF. Join
BE and DC, DG and FE,
by lines intersecting in the
points H and K. Draw a
line through H and K till
TANGENTS AND TANGENTIAL ARCS.
59
it meets the circumference in L and M. Join AL and AM,
which will be the tangents required.
89. In a given angle CAB, to inscribe a circle wliich shall pass
through a given point D,
Bisect the angle CAB,
by the line AE (Prob. 62).
Take any convenient
point F in AE, and from
F draw the line FG per
pendicular to AC (Prob.
7). With F as centre,
and FG as radius, draw a
circle. Join DA, cutting
the circle in H. From
the given point D draw
DK parallel to HF. With
K as centre, and KD as
radius, draw the required
circle.
Fig. 127.
90. To draw a circle which shall pass through the given point
A and touch a given line BC in D.
At the given point D
erect the Hne DE perpen
dicular to BC (Prob. 5),
and join AD. At point A
construct an angle DAE
equal to the angle ADE
(Prob. 11). AE intersects
DE in E. With E as
centre, and ED as radius,
draw the required circle.
6o
GEOMETRICAL DRAWING AND DESIGN.
91. To draw a circle which shall pass through the two given
points A and B and touch the given line CD.
Join the two given
points AB, and produce
the line till it meets the
given line CD produced
in E. On AE describe
the semicircle EFA ; at B
draw BF perpendicular to
AE. From E along the
line ED, set off EG equal,
to EF. Then G, B, A
are three points in thd
required circle, which can
be drawn as required
(Prob. 33).
92. To draw four equal circles, with radius equal to given line
E, to touch two given lines AB and CD, which are not parallel.
Bisect the two adjacent,
angles by the lines FGj
and HK. Draw the lines
LM and NO parallel tc
the given line CD, at a
distance from it equal tc
the given radius E (Prob.
4). Where these lines
intersect the bisectors, we
get the points S, R, T, and
P. With the points R, S,
T, and P as centres, and
with a radius equal to E,
draw the four circles re
quired.
TANGENTS AND TANGENTIAL ARCS.
6i
93. To draw an inscribed and an escribed circle, tangential to
three given straight lines, forming a triangle ABC.
Note. — An escribed circle is also called an excircle.
Bisect the angles BAG and ACB
by lines intersecting in F. From
F draw the line FH perpendicular
to AD. With F as centre, and
FH as radius, draw the inscribed
circle required. Bisect the angle
BCD by a line cutting the line
AG in E. Draw the line EK per
pendicular to AL. With E as
centre, and EK as radius, draw the
escribed circle required.
94. A principle of inscribed and escribed circles.
If a triangle ABC be taken, and
AF, BD, and CE be lines drawn
from the three angles perpendicular
to the opposite sides, they will all
intersect at the point H. Join the
points D, E, F. This will form a
triangle of which H is the "in
centre," being the centre of the
inscribed circle. The centres of
the escribed circles will be the
points A, B, and C. The radii of
the circles are found by drawing
lines from the centres perpendicular
to the sides of the triangle produced (Prob. 7),
line BK.
Fig. 132.
as the
62
GEOMETRICAL DRAWING AND DESIGN.
95.
To draw two circles tangential to three given straight lines,
two of which are parallel.
Let AB and CD be the two given
parallel lines, and let the third line
intersect them in E arid F. Bisect
the four angles AEF, BEF, CFE,
and DFE by lines meeting at H
and L. From H draw the line HM
perpendicular to CD (Prob. 7).
With H and L as centres, and a
radius equal to HM, draw the two
required circles.
Note. — A line joining H and L will be parallel to the two
given lines AB and CD.
Fig 133
96. To draw two circles tangential to three given straight lines,
none of which are parallel ; the third line to be drawn to cut
the other two.
Let AB, CD, and EF be the three given lines. Bisect the four
j3 angles AFE, BFE, CEF,
and DEF by lines meeting
at H and L. From H and
L draw the lines HM and
LN perpendicular to CD
(Prob. 7). With H as
centre, and radius HM,
draw a circle ; with L as
centre, and LN as radius,
draw the other circle re
quired.
Note. — A line pro
duced through the points
H and L would be the
bisector of the angle formed by producing the Hues AB
and CD.
TANGENTS AND TANGENTIAL ARCS.
63
97. To draw DIRECT COMMON TANGENTS tO tWO given
circles of unequal radii.
Let AC be the radius of one circle, and BD of the other.
Join the centres A and B. ,. .^
From the centre A draw ~^!^^^^^=="r?==~— ____^
a circle with a radius /^ ^j N^ '~~y>:^ — J
= ACBD. Bisect the line / (^\~'\~Zr~~r~Jr~'\\
AB at E (Prob. i). From I \^_'J:,.y^'^^^ J
E, with radius ExA., draw a \ ^, y _— — 7'^'^^*
circle cutting the circle FKG ___^^^^=1p\
in the points F and G. Join ' ''
FB and GB. From F and B ^' ^35
draw the lines FO and BR perpendicular to FB (Prob. 7) ;
and from the points G and B draw the Hnes GP and BS
perpendicular to GB. Draw the line HL through the points
O and R, and the line MN through the points P and S ; these
will be the tangents required.
98. To draw TRANSVERSE COMMON TANGENTS tO tWO given
circles of unequal radii.
Let A and B be the centres of the given circles, and AC and
BD their radii. Join AB.
With A as centre, and a
radius = AC + BD, draw a
circle. Bisect the line AB
in E (Prob. i). With E as
centre, and a radius equal
to EA, draw a circle cutting
the circle FKG in the points
F and G. Join AF and AG,
cutting ths given circle PCO
in O and P. Join FB and GB. From B draw the line BS per
pendicular to FB (Prob. 7), also the line BR perpendicular to
GB. Draw the Hne HL through the points O and S, and the
hne MN through the points P and R ; these will be the
tangents required.
Fig. 1,6.
64
GEOMETRICAL DRAWING AND DESIGN.
Tangential Circles and Arcs.
99. Showing the principle of tangential circles.
One circle can touch another circle either internally or ex
ternally, and any number of circles can be drawn to touch a
given line, as well as each
other, in the same point. For
instance take the point C on
the given line AB. All circles
that touch a given line in the
same point touch each other at
that point ; and all their centres
will be on a line perpendicular
to the given hne.
If they are on the opposite
sides of the given line, they will
touch externally ; and if on the
same side, will touch internally.
If they are on the same side
of the line, one circle will be
entirely within the other.
If their centres F and E are on the same side of the given
line AB, the distance between them is equal to the difference
of their radii ; but if their centres E and D are on the opposite
sides of the given line, the distance between them is equal to the
l' sum of their radii.
__i.^_ The point of contact
^ '' 'q " '^ >. ^ can always be found by
joining their centres.
100. To draw four equal
circles with radius
equal to given line D,
with their centres on a
given line AB ; two to
touch externally and
two internally a given
_ circle, whose centre is
I C and radius CG.
13 ' With centre C, and
Fig. 138.
TANGENTS AND TANGENTIAL ARCS.
65
radius equal to the sum of the radii, i.e. CG + D, draw a circle
cutting the given Hne AB in H and N. With centre C, and
radius equal to the difference of the radii, i.e. CO  D, draw a
circle cutting the given line AB in L and M. With H, L, M,
and N as centres, and radius equal to D, draw the four circles
required.
101. To draw four equal
circles, with radius
equal to given line D,
with their centres on a
given arc AB ; two to
touch externally and
two internally a given
circle, whose centre is
C and radius CG.
The construction of this
problem is word for word
the same as the last, the
only difference being the
words given arc instead of
given line.
Fig. 139.
102 To describe a circle tan
gential to and including
two given equal circles A
and B, and touching one of
them in a given point C.
Find the centres of the two
given circles D and E, and
join them (Prob. 32). Join
CD. From C draw the line
CK parallel to DE, meeting
the given circle B in K. Join
KE, and produce it till it
meets CD produced in F.
With F as centre, and radius
FC, draw the required circle.
Fig. 40.
66
GEOMETRICAL DRAWING AND DESIGN.
103. To describe a circle tangential to and including two unequal
given circles A and B,
Fig. 141.
and touching one of them
in a given point C.
Find the centres D and
E. Join CE. Cut off from
CE, CH equal in length to
the radius of the smaller
circle. Join DH. Produce
CE. At D construct the
angle HDF equal to the
angle DHF(Prob. 11). DF
meets the line CE produced
in F. With F as centre,
and FC as radius, draw the
circle required.
104. To draw the arc of a circle having a radius of l: inches, which
. p shall be tangential to two
\ given unequal circles A and
B and include them.
Note. — The diameters and
distance between the circles
must not be greater than 2^
inches.
Find the centres D and E
(Prob. 32), and produce a hne
through them indefinitely in
both directions, cutting the
circles in K and L. From the
points K and L on this line, set
off KF and LH equal to the
radius of the required arc, viz.
I J inches. With D as centre,
and a radius equal to DF, draw
an arc at M ; and with E as
centre, and EH as radius, draw
Fig. 142.
an arc intersecting the other arc at M. From M draw the
line MD, and produce it till it meets the circumference of the
TANGENTS AND TANGENTIAL ARCS.
67
larger circle in C. With M as centre, and MC as radius, draw
the required arc.
105. To inscribe in a segment of a circle, whose centre is E, two
given equal circles with a radius equal to line D.
From any radius EF cut off
FL = D, and describe a circle with
radius EL. Draw the line KL parallel
to the base of the segment, and at a
distance equal to given radius D from
it (Prob. 4). Join EL and produce
it till it meets the circumference in
F. With L and K as centres, and
LF as radius, draw the two required
circles.
106. In a given sector of a circle ABC,
to inscribe a circle tangential to
it.
Bisect the angle ACB by the line CD
(Prob. 12). At D draw a tangent HL
(Prob. 84) to meet the sides of the
sector produced. Bisect the angle
CLH by a line cutting CD in E.
With E as centre, and ED as radius,
draw the circle required. ^
Fig. 144.
107. Draw a circle having a radius of j of an inch tangential to
two given unequal circles A and B externally.
Note. — The circles must not be
more than h inch apart.
Find the centres D and E of the
given circles (Prob. 32). From centre
D, with the sum of the radii, z.e.
D K + 1 of an inch, draw an arc at H ;
and from centre E, with the sum of the
other radii, z>. EL + j of an inch, draw
another arc at H . With H as centre, and
radius HK, draw the circle required.
Fig. 145.
68
GEOMETRICAL DRAWING AND DESIGN.
108. To draw the arc of a circle tangential to two given unequal
circles A and B externally, and touch
ing one of them in a given point C.
Find the centres D and E of the
given circles (Prob. 32). Join CE, and
produce it indefinitely. Set off from
C, on CE produced, CH equal to the
radius of the larger given circle. Join
DH. At D construct an angle HDF,
equal to the angle FHD, to meet EC
produced in F. With F as centre, and
radius FC, draw the arc required.
109. To draw a circle, with a radius equal to given line C, tangential
to two given unequal circles A and B,
to touch A externally and B internally.
Note. — The given radius must be
greater than half the diameter of the
enclosed circle and the distance
between the circles.
Find the centres D and E of the two
given circles (Prob. 32). From centre
D, with radius, the sum i.e. DH + C,
describe an arc at F ; and with E as
centre and radius the difference, i.e.
C — EK, draw another arc cutting the
FD and FE. With F as centre, and
Fig. 147.
Other at point F. Join
FH as radius, draw the circle required.
110.
To draw a circle of f of an inch radius tangential to the
given line AB and the given circle CDE.
Note. — The circle must be less than
I inch from the line. Draw the line
KL parallel to AB, and f of an inch
from it. Find the centre F of the
given circle (Prob. 32), and with the
sumradius of the two circles draw the
arc HM cutting KL in M. Join FM.
With M as centre, and MN as radius,
draw the required circle.
TANGENTS AND TANGENTIAL ARCS. 69
EXERCISES.
1. Draw a circle 1.7" in diameter; at any point in its circumference,
draw a tangent.
2. Draw a circle 1.25" in radius ; from a point one inch outside the
circle, draw a tangent to it.
3. Draw a circle i" in diameter ; from any point outside the circle,
draw two tangents without using the centre.
4. Draw two lines, enclosing an angle of 45° ; draw a circle i" in
diameter, tangential to these lines.
5. At any point in the arc of a circle draw a tangent, without using
the centre.
6. Draw two circles of 2" and i" radii, with their centres 3" apart ;
draw transverse common tangents to them.
7. Draw two circles of 1.7" and i" radii, with their centres 2.75"
apart ; draw direct common tangents to them.
8. Draw two lines at an angle of 30° with each other, and a third
line cutting them both at any convenient angle ; draw two circles
tangential to all the three lines.
9. Construct a triangle with sides 2.25", I.S", and 1.25"; draw an
inscribed and three escribed circles tangential to the lines forming the
sides.
10. Draw two lines enclosing an angle of 45° ; fix a point in any
convenient position between these two lines, and draw a circle that
shall pass through this point and be tangential to the two lines.
11. Draw a circle 1.25" in diameter, and half an inch from it draw
a straight line ; draw a circle of §" radius, tangential to both the
circle and the line.
12. Draw two circles of i" and V' radius, with their centres 2h"
apart ; draw another circle tangential to both externally.
13. Draw two circles 1.50" and i" in diameter, their centres to be
2.25" apart; draw another circle 3^" in diameter, touching the larger
circle externally, and the smaller one internally.
14. Draw two lines AB, AC, making an angle of 25° at A. Describe
a circle off" radius touching AB and having its centre on AC. From
A draw a second tangent to the circle, marking clearly the point of
contact. (April, '96.)
15. Draw a line AB, 2" long. Describe a circle of f " radius touching
AB at A, and another of i" radius touching AB at B. Draw a second
line which shall touch both circles, showing clearly the points of
contact. (June, '97.)
16. Describe two circles, each of h" radius, touching each other at a
point A. Find a point B on one of the circles, t" from A. Describe a
third circle, touching both the others and passing through B. (April, '98. )
7o
GEOMETRICAL DRAWING AND DESIGN.
17. Construct a square of 2" side. In the centre of the square place
a circle of ^" radius. Then describe four other circles, each touching
the first circle and two sides of the square. (June, '00. )
18. Draw the "cyma recta" moulding shown, (Fig. 149), adhering to
the given dimensions. The curve is composed of two quartercircles
of equal radii, tangential to one another and to the lines AB and CD
respectively. (April, '96.)
Fig. i4q.
Fig. 151.
19. Draw the "Scotia" moulding shown (Fig. 150). The curve is
made up of two quartercircles of 1" and " radius respectively.
(May, '97.)
20. Draw the "rosette" shown (Fig. 151), according to the figured
dimensions. (June, '97).
21. Draw the "ogee" arch shown (Fig. 152) to a scale of 2' to i".
The arcs are all of 2' radius. The methods of finding the centres and
points of contact must be clearly shown. (April, '99.)
22. Draw the moulding shown (Fig. 153), adhering strictly to the
figured dimensions. The arc of ^" radius is a quadrant. (April, '98.)
23. Draw the "cyma recta" moulding shown in the diagram (Fig.
154), using the figured dimensions. The curve is composed of two
equal tangential arcs each of f" radius. (June, '98.)
24. Describe a circle touching the given circle at
T (Fig. 155) and passing through the point P (twice
size of figure). (June, '99.)
25. Describe the four given circles, using the
figured radii (Fig. 156). The necessary construction
lines and points of contact must be clearly shown. (June, '00.)
^ig 155
TANGENTS AND TANGENTIAL ARCS.
71
26. Draw the "three centred" arch shown (Fig. 157). The two
lower arcs are of i" radius, and the upper arc is of 2^" radius. (April, '00. )
Fig. 156.
Fig. 157
Fig. 158.
27. Draw the "four centred" arch shown (Fig. 158), adhering to the
figured dimensions. (June, '97.)
28. Draw the figure shown, (Fig. 159), making tlie sides of the square
21" long. (April, '96. )
*^
^
—  3^"  >
iimi
i^^O^
Fig. 161.
Fig. 160.
Fig. 159
29. Draw the figure shown (Fig. 160), adhering strictly to the figured
dimensions. (May, '97.)
30. Draw the figure shown (Fig. 161), according to the figured
dimensions. (June, '97.)
31. Draw the given figure (Fig. 162) making the side of the square
2^" and the radii of all the arcs f". (April, '99,)
W
■£k
2^'
Fig. 162.
Fig. 164.
32, Draw the figure shown (Fig. 163), using the figured dimensions.
(April, '98.)
33. Copy the given figure (Fig. 164), using the figured dimensions.
(June, '00.)
CHAPTER VII.
PROPORTIONAL LINES.
Proportional lines may be illustrated by the example of simple
proportion in arithmetic, in which we have four terms, e.g.
2:4::5 : lo
The relationship or ratio between the first two terms with
regard to magnitude is the same as that between the second two,
e.g. as 2 is to 4, so is 5 to ID ; therefore these four numbers
are said to be in proportion.
The first and fourth terms are called
the extremes, and the second and third the
means.
The product of the extremes equals the
product of the means, e.g. 2x10 = 4x5.
So, the first three terms being given, we
can find the fourth. If we divide the pro
duct of the means by the first extreme we
get the fourth proportional, e.g.  — = 10.
Almost all geometrical questions on
proportion are based on the following
theorems : —
Take any triangle ABC, and draw any
line DE parallel to one side, then —
CD :DA::CE:EB
CD : CE : : CA : CB
CE:ED::CB:BA
CD : DE : : CA : AB.
There are five varieties of proportional lines, viz. —
Greater fourth proportional.
Less fourth proportional.
Greater third proportional.
Less third proportionaL
Mean proportionaL
PROPORTIONAL LINES.
73
If the quantities be so arranged that the second term is
greater than the first, — as 4 : 6 : : 8 :.v, — the last term is called
the greater fourth proportional.
If the terms are arranged so that the second term is less than
the one preceding, — as 8 : 6 : : 4 : .r, — the last or unknown term
is called the less fourth proportional.
When the two means are represented by the same number, —
thus 4 : 6 : : 6 : .r, — the answer or x is called the third propor
tional.
The third proportional is found by dividing the square of the
second by the first, e.£: —
62 6x6
— or = 9.
4 4
If the terms are placed so that the larger number is repeated,
—thus 4:6::6:.i', — the last term is called the greater third
proportional ; but if the terms are arranged so that the smaller
number is repeated,— as 6 : 4 : : 4 : .r, — the result is called the
less third proportional.
The mean proportional between any two numbers is found by
extracting the square root of their product, — e.g: 4x9 = 36 ; the
square root of 36 = 6, which is the mean proportional.
111. To find a fourth proportional to three given lines A, B,
and C. THE GREATER FOURTH PROPORTIONAL.
Draw EH equal to
given line C, and EF, at
any angle with it, equal to
given line B. Join FH,
and produce EH to D,
making ED equal to given
line A. Draw DK parallel
to FH till it meets EF
produced in K (Prob. 3).
Then KE will be the
greater fourth proportional
to the lines A, B, and C,
i.e. C : B : : A : KE, e.g.
if C = 6feet, B = 8 feet, A =
A.
B: .
C' '
Fig. 166.
■ 12 feet, then KE=^ 16 feet
74
GEOMETRICAL DRAWING AND DESIGN.
112. To find a fourth proportional to tliree given lines A, B, and C.
THE LESS FOURTH PROPORTIONAL.
Draw the line DE
equal to given line A,
and EF, at any angle
with it, equal to given
line B. Join FD. From
E, along ED, set off
EG equal to given line
C. From G draw GH
parallel to FD (Prob.
3). Then HE is the
less fourth proportional to
the given lines A, B, and
C, /.^. A : B : : C : HE.
Fig. 167.
113. To find a third proportional between two given lines A and B.
THE GREATER THIRD PROPORTIONAL.
Draw CD equal to given
line A, and CE, at any
angle with it, equal to
given line B. Join DE.
With C as centre, and CD
as radius, draw the arc. DG
to meet CE produced in
G. From G draw the line
GF parallel to DE till it
meets CD produced in F
^ (Prob. 3). Then CF is the
greater third proportional
' to the given lines A and B,
Fig. 168. I.e. B : A : : A : CF.
114, To find a third proportional between two given lines A and B.
THE LESS THIRD PROPORTIONAL.
Draw CD equal to the given line A, and CE, at any
PROPORTION.
75
angle with it, equal to
given line B. Join DE.
From C as centre, and
with radius CE, draw the
arc EF cutting CD in F.
Draw FG parallel tc DE
(Prob. 3). Then CG is
the less third proportional
to the given Hnes A and B,
z\e.A:B::B: CG.
\ 1
\
^v^
G
C
Fig. 169.
115. To find the MEAN PROPORTIONAL between two given lines
AB and CD.
Produce the given line AB
to E, making AE equal to
the given line CD. Bisect /^
the line EB in H (Prob. i).
From H as centre, and with /
radius HB, draw the semi /
circle EKB. At A draw the /
line AK perpendicular to EB, ~
cutting the semicircle in K
(Prob. 5). Then AK is the 1
mean proportional to the given ^
lines AB and CD, e.o. if
AB = 9 feet and CD 4 feet, then AK:
H
A
Fie.
6 feet.
116.
To divide a line in medial section, i.e. into EXTREME and
MEAN proportion.
Let AB be the line. At A
draw AC perpendicular to
AB, and equal to it. Bisect
AC in D. With D as centre,
and DB as radius, draw the
arc cutting CA produced in
E. With A as centre, and
AE as radius, draw the arc
cutting AB in F. Then
AB : AF : : AF : BF.
B
76
GEOMETRICAL DRAWING AND DESIGN.
117. To divide any straight line AB in the point C, so that
AC : CB : : 3 : 4.
At A draw the line AD of
indefinite length, and at any
angle to AB. From A, along
AD, mark off seven equal
distances of any convenient
length. Join 7B. At 3 draw
the line 3C parallel to 7B
(Prob. 3). Then
AC : CB : : 3 : 4,
or, AB : AC : : 7 : 3.
118. To divide a line proportionally to a given divided line.
Let AB be a given divided
line, and GH a line which is
to be divided proportionally
to AB. Join GA and HB,
and produce them to meet in
C ; join C to the divisions
in AB and produce them.
These lines will divide GH
as required. Divide BC into
four equal parts, and draw
the lines D, E, and F parallel
to AB ; join the divisions on
AB with C, then the divisions
on the lines D, E, and F will
represent respectively the
proportions of , ^, and j
of the divisions on the given
line AB.
/ / ' ,
f • >
^ / / i — i
\^ — / — f '.
_/ / hJ
PROPORTIONAL LINES.
77
119. To construct a triangle on a given line AB, so that the
three angles may he in the proportion of 2 : 3 : 4.
From B, with any radius, describe a semicircle and divide it
into nine equal parts (Prob. 48). Draw the lines B4 and B7.
Then the three angles 7BC,
4B7, and 4BA are in the
proportion of 2 : 3 : 4. The
sum of the three angles are
equal to two right angles,
because a semicircle con
tains 180° ; so they must
be the three angles of a
triangle, because the three
angles of any triangle are
together equal to two right angles. From A draw the line
AD parallel to B7 till it meets B4 produced in D (Prob. 3).
Then ABD is the triangle required.
120. This problem illustrates an important principle in proportion.
Take a triangle ABC, the sides of which shall bear a
certain ratio. For example,
let AB : BC as 2 : i. Pro D
duce AB to D, and bisect the / ^,
angles ABC and DBC by
lines meeting AQ and AC
produced in H and E.
Bisect the line HE in K
(Prob. i). With K as centre,
and KE as radius, draw the
circle EBH. Now, if we
take any point M in this
circle, and join MA and MC,
we shall find that they bear
the same ratio as the lines
AB and BC. In the example
given MA : MC as 2:1. The same result would be obtained
from any point in the circle.
78
GEOMETRICAL DRAWING AND DESIGN.
121. To divide a right angle into five equal parts.
Let ABC be the right
angle. Divide BC in D, so
that BC : BD : : BD : DC
(Prob. 1 1 6). With C as
centre, and CB as radius,
describe the arc BE ; and
with B as centre, and BD as
radius, describe the quadrant
DF, cutting BE in E FE is
onefifth of the quadrant FD.
Arcs equal to it set off on
FD will divide it into five
equal parts.
122
— ^.E
To find the Arithmetic, the Geometric, and the Harmonic means
between two given lines AB and BC.
Bisect AC in D (Prob. i).
With D as centre, and radius
DA, draw the semicircle
^ AEC. At B draw BE per
'^ pendicular to AC (Prob. 5).
\ Join DE. From B draw BF
I perpendicular to DE. AD
D B C. is the Arithmetic, BE the
Geometric, and EF the
177.
Harmonic mean between the two lines as required.
Fig. 178.
123. Taking the given line AB as the unit ; find
lines representing \^ and sfs.
Draw AC perpendicular to AB, and of the
same length (Prob. 5). Join i:B. CB = n/2.
Draw CD perpendicular to CB, and equal
in length to AB and AC. Join DB. DB = v'3.
If AB is the edge of a cube, CB is the
diagonal of its face, and DB the diagonal of
the cube, which are therefore to one another
as I : \/2 : \/3.
PROPORTIONAL LINES.
79
Table of Foreign Road Measures and their
Equivalents in English Yards.
English Yards.
Austria,   •
mile
8297
Bavaria, 
))
8059
Belgium,    
kilometre
1094
Berne,    
league
5770
China, . =  
li
609
Denmark,
mile
8238
England,   = 
55
1760
France,    
kilometre
1094
Germany,
mile
8101
Greece,    
))
1640
Holland,    
5»
1094
India (Bengal),
coss
2000
Italy,    
mile
2025
Netherlands, 
kilometre
1094
Norway,    
mile
12,182
Persia, .  . 
parasang
6076
Portugal,    
mile
2250
Prussia,    
5>
8238
Russia,    
verst
I167
Siam, ... 
roeneng
4204
Spain, ... 
mile
1522
Sweden, .   
55
11,690
Turkey,    
berri
1827
EXERCISES.
1. Draw three lines 1.25", 2.3", and 2.75" respectively, and find
their greater fourth proportional.
2. Draw two lines 2.7" and 1.5" in length, and find their less third
proportional.
3. Draw a line 2.5" in length, and produce it so that its extra length
shall be in proportion to its original length as 3 : 5.
4. Draw two line's 25" and " in length, and find their mean
proportional.
5. Construct a triangle on a base 3.25" in length, so that its three
angles are in the proportion of 3, 4, and 5.
8o GEOMETRICAL DRAWING AND DESIGN.
6. Divide a line 3.4" in Kngth in extreme and mean proportion.
7. Divide a line 2.75" in length so that one part is in proportion to
the other as 2 : 4.
8. Draw two lines 1.25" and 2.3" respectively, and find their greater
third proportional.
9. Draw three lines 3", 2^", and i" in length, and find their less
fourth proportional.
10. The base of a triangle is l '6^" long, and the angles at the base
are 88° and 53°. Construct the triangle, and find a fourth proportional
less to the three sides. Measure and write down the length of this line.
(The angles should be found from the protractor or scale of chords.)
(April, '01.)
CHAPTER VIII.
PLAIN SCALES, COMPARATIVE SCALES, AND
DIAGONAL SCALES.
On a drawing representing a piece of machinery the scale is
written thus : Scale j full size. From this we know that every
inch on the drawing represents 4 inches on the actual machine,
so the relation between any part represented on the drawing and
a corresponding part in the real object is as i : 4 or j. This is
called the representative fraction.
A drawing representing a building has drawn upon it a scale;
e.g. — Scale \ of an inch to a foot. Onequarter of an inch is
contained fortyeight times in i foot, so the R.F. is ^^.
On a large drawing showing a district the scale is written
thus : R.F. yyVo As there are 1760 yards to a mile, it is
evident that every 3 feet on the drawing is equal to i mile
on the land represented. This, of course, is a very large
scale.
Our Ordnance Survey Office publishes a map of 25 inches to a mile,
which is useful for small districts or estates ; one of 6 inches to a mile,
useful for maps of parishes ; and one of I inch to a mile, useful for
general purposes.
The R.F. for the last would be g^eo
mile, yards. feet. inches.
I = 1760=5280 = 63,360.
F
82
GEOMETRICAL DRAWING AND DESIGN.
124.
To construct a scale 4 inches long, showing inches and
tenths of an inch. Fig. 179.
Draw a line 4 inches long, and
divide it into four equal parts, each
of which will be i inch. At the end
of the first inch mark the zero point,
and from this point mark the inches
to the right i, 2, 3. These are called
primary divisions, and the amount
by which they increase is called the
value of the scale length.
The division left of the zero
point has to be divided into ten
equal parts. The best way to do
this is to take a piece of paper and
set off along its edge ten equal
divisions of any convenient size
(Fig. 179). Produce the perpendi
cular marking the division at the
zero point, and arrange this piece
of paper so as to fit in exactly
between the end of the division and
this perpendicular line. If we now
draw lines parallel to the perpendi
cular at the zero point, they will
divide the inch into ten equal parts.
These are called the secondary
divisions ; they have the same zero
point as the primary, and their
numbers increase from this poini
by the value of their scale length.
This scale will measure inches
to one decimal place. Supposing
we wish to measure 3.6 inches,
that is 3 primary and 6 secondary
divisions. Place one point of the
\
.\X V
 ro ^
1
Co
Fig. 179.
dividers on division 3 of the primary parts, and open them
till the other point reaches the division (^ of the secondary
divisions.
PLAIN, COMPARATIVE, AND DIAGONAL SCALES. 8
125.
<T>
00
cs
lO'
HI?
^
to
CVJ
To construct a scale of g^g, or 1 incli to equal 3 feet
Fig. i8o.
Number of feet to be repre
sented may be assumed at
pleasure, say I2 feet.
36 : 12 : : 12 :;r,
I2X 12
whence ,r:
4 inches.
Fig. 180.
36
Draw a Hne 4 inches long,
and mark off each inch. Trisect
each of these divisions by a piece
of paper, as shown in Fig. 179.
We now have the total length
divided into twelve ecjual parts.
At the end of the first division
*§ mark the zero point, and from
^•^ this point towards the right,
^ figure the primary divisions i,
2, 3, etc. : these will represent
feet. To the left of the zero
point mark oflf twelve divisions :
these will represent inches.
In this scale the numbers
increase on each side of the
zero point by unity.
This scale will measure feet
and inches.
126. To construct a scale with the
R.F. os, or 1 inch to equal
8 yards ; to measure 40 yards.
Fig. 18 T.
288 : 40 : : 36 :;r,
36x40
whence x
288
= 5 mches.
O
Draw a line 5 inches in length, and divide it
into four equal parts, i.e. 1.25 for each division.
At the end of the first division mark the zero
point. As each of the primary divisions is
O
00
O.
Fig. 181
I
84
GEOMETRICAL DRAWING AND DESIGN.
ro
C\]
O
c\J
CD
Ik
Fig. 182.
^
^
equal to 10 yards, we must
figure them from the zero
point to the right 10, 20,
and 30 yards.
The division to the left
of the zero point we divide
into ten equal divisions.
Each of these secondary
divisions represents i
yard.
127. To construct a scale,
RF. 6 356 o> or 1 iiich
to a mile ; to measure
5 miles. Fig. 182.
If I mile=i inch, 5
miles = 5 inches.
Draw a line 5 inches
long. Mark off each inch.
At the end of the first
division mark the zero
point, and number the
primary divisions to the
right I, 2, 3, 4 miles.
Divide the division to the
left of the zero point into
eight equal parts : these
secondary divisions repre
sent furlongs.
Comparative Scales.
o
0
0J
On an old French map
a scale of leagues is ^
shown, as Fig. 183. Upon
measuring this scale with qo
an English scale, 30
leagues are found to coin oo
cide with 4 inches.
o
00
I
o __
CM
o ■
CO
00
O 
Fig. i»3.
PLAIN, COMPARATIVE, AND DIAGONAL SCALES. 85
128. To construct
Scale to
A French league =
French
leagues, miles,
4262.84
a comparative scale of English miles
measure 80 miles. Fig. 183.
4262.84 English yards.
English O
30 =
1760
X 30
4262.84 X 30
1 760
whence
4 X 80 X 1 760
: 80 : : 4 : x
4.4 inches nearly.
4262.84x30
Draw a line of this length, and place
the zero point at the lefthand end and
80 at the other extremity. Divide this
line into eight equal divisions : each
of these primary divisions will repre
sent 10 miles.
For the secondary divisions, set off
one of the primary divisions to the left
of the zero point, and divide it into ten
equal divisions : each of these will
represent i mile. The representative
fraction of both the French and English
scales will of course be the same.
On a Russian map a scale of versts
is shown, as Fig. 184, by measuring
which by an English scale
120 versts = 4 inches.
129. To construct a comparative scale of
English miles Scale to measure 80
miles. Fig. 184.
A Russian verst=ii67 English yards.
1 167 X 120
120 versts =
1 167 X 120
1760
miles.
1760
80 : : 4 : X,
o— 
I
o
00
whence
4 X 80 X 1 760
1167x120
= 4 inches nearly.
R
o
o
o
CO
o
CM
o
CVJ
to
00
o_
^
Fig. 184.
86 GEOMETRICAL DRAWING AND DESIGN.
Draw a line of this length, and divide it into eight equal
divisions : each of these primary divisions will represent lo miles.
Place the zero point at the lefthand end of the line, and figure
the divisions towards the right lo, 20, 30, etc. Set off one of
the primary divisions to the left of the zero point, and divide it
into ten equal divisions : each of these will represent i mile.
Diagonal Scales.
In the preceding scales we have only primary and secondary
divisions, and if we wish to measure a fractional proportion of a
secondary division, we cannot do it with any accuracy ; but by
means of a diagonal scale we are enabled to measure hundredths
of primary divisions, as will be seen from the following scale.
130. To construct a diagonal scale 3 inches long, to measure inches,
tenths of inches, and hundredths of inches. Fig. 185a.
Draw a rectangle ABDC 6 inches long and about ij inches
wide, and divide it into six equal parts. At the end of the first
I" I I I I I
Fig. i«5a.
Fig. 185b.
N.B. — These two figures are half the size described in the text, and should be
drawn full size by the student.
division from A fix the zero point, and to the right of this
figure each division i, 2, 3, 4, and 5. Divide AC into ten
equal parts, and figure them from A towards C, then draw lines
parallel to AB from one end of the scale to the other. Divide
PLAIN, COMPARATIVE, AND DIAGONAL SCALES. 87
Ao into ten equal divisions, and figure them from o towards A.
Join 9 to C, and from each of the other divisions between A
and o draw hnes parallel to 9C. Note. — The divisions between
o and B are primary, between o and A secondary and between
A and C tertiary.
To take off from this scale a measurement equal to 2.73
inches, we place one point of the dividers on the primary
division figured 2, and the other on the secondary division
figured 7, but both points must be on the line that is figured 3
on AC. The points are marked by a small circle on the scale.
To take off 3.17 inches, place one point of the dividers on
the primary division 3, and the other on the intersection between
the secondary division i and the line 7 on AC. These points
are shown by two crosses on the scale.
131. To construct a diagonal scale showing miles, furlongs, and
chains, to show 6 miles, R.F, = g3.^gQ. P^ig. j85b.
I mile = 8 furlongs.
T furlong=io chains.
The length of scale = g3^go of 6 miles = 6 inches.
In this scale there will be six primary, eight secondary, and
ten tertiary divisions.
Construct a rectangle ABDC 6 inches long and about ij
inches wide, and divide it into six primary divisions. Place the
zero point o at the end of the first division from A, and divide
Ao into eight secondary divisions, figured from o towards A.
Divide AC into ten equal divisions, and figure them from A to
C. Join the secondary division figured 7 to the point C, and
from each of the other secondary divisions draw lines parallel
to 7C, thus completing the scale.
To take off from this scale 2 miles, 5 furlongs, and 7 chains,
take one point on the primary division 2, and the other where
the line from the secondary division 5 intersects the division
7 on AC. These two points are marked by two small circles on
the scale.
To take off 3 miles, 2 furlongs, and 3 chains, one point will be
on the primary division 3, and the other where the secondary
division 2 intersects the tertiary division 3. These points are
marked by two small crosses on the scale.
88
GEOMETRICAL DRAWING AND DESIGN.
132. To take off any number to three places of figures from
a diagonal scale.
On the parallel indicated by the third figure, measure from the
diagonal indicated by the second figure
to the vertical line indicated by the first.
DISTANCE
61
lOLOMtrJlES
Fig. 186.
Comparative Diagonal Scales
useful for transferring the
value of quantities in one
measure to another. To
make the scales less cum
bersome they are so
arranged that in some
cases the number to be
taken off must be halved.
With this proviso, any
quantity may be converted
from one scale to another.
The number expressing
the quantity in one unit is
taken off on the scale for
that unit, and the number
expressing it in the other
unit is at once read off on
the parallel scale.
For example, a length
of 638 miles. Its half,
319 miles, corresponds to
5 1 3 kilometres, so that 638
miles corresponds to 1026
kilometres.
The diagonal scale
generally found in instru
mentboxes is shown in
Fig. 187.
diagonal scales. In one,
are very
3) y>
«N
7>
>o
f
<t>
^■■U
■■■«»

■ f
4 .
■ rt
: , II
1 ty
_
~

CD
 T
 i  ■
•r"
:i:,:
:::
!:::"
eoo >■
el
Fig. 187.
It consists of two diagonal scales. In one, the distance
between the primary divisions is half an inch, and in the other
a quarter of an inch.
PLAIN, COMPARATIVE, AND DIAGONAL SCALES. 89
There is a small margin on each side of the scale for figures :
on one side the half inches are figured, and the quarter inches
on the other.
One primary division at each end is divided into ten secondary
divisions, and there are ten tertiary divisions drawn from one
end of the scale to the other.
The primary divisions bemg taken for units, to set off the
numbers 5.36 by the diagonal scale. This measurement is
shown by two crosses on the scale.
If we reckon the primary divisions to stand for tens, the
dimension would have one place of decimals, e.g. to take off 36.4
from the diagonal scale. These points are shown on the scale
by two small circles.
The primary divisions being hundreds, to take off 227. This
dimension is shown on the scale by two small squares.
Proportional Scales.
These are used for enlarging or reducing a drawing in a given
proportion : three varieties are here illustrated.
The simplest form is that shown in Fig. 188. Suppose we wish
to enlarge a drawing in the proportion of 3 : i.
Draw the line AB of convenient
size, to suit the measurements on
the drawing, and produce it to C ;
make BC onethird of AB. On AB
erect the perpendicular BD any
length, and join AD and DC.
Divide BD into any number of
equal parts, and draw lines parallel
to AB. These lines are simply a
guide to enable the measurements
to be made parallel to the base —
e.g. on placing a measurement from Fig. 188.
the original drawing on the scale
we find it occupies the position of ef: the distance between e
and g will then give the length of the measurement to the
enlarged scale, i.e. in the proportion of 3 : i. We proceed in the
same manner with every measurement we wish to enlarge.
D
\
/
~~\
./
e V
/ :
/
\
A
i c
@
■■•••.
^. /
F ty ^
/
 — ;>%.
/rt
/9
I
V'
90 GEOMETRICAL DRAWING AND DESIGN.
Should we wish to reduce a drawing in the same proportion,
viz. I : 3, the original measurements would be placed on the
lefthand side of the scale, and the required proportion taken
from the righthand side.
In Fig. 189 we have a series of measurements — Az', Kh^ Ag^
etc. — which we wish to enlarge, say in the proportion of 3 : 2.
^ Draw the line AB any convenient
length to suit size of drawing.
From B draw BD perpendicular
to AB. Produce AB to C, and
make BC equal to half of AB.
With centre A, and radius AC,
draw an arc till it meets BD in D ;
and join DA. From each of the
points, /, //, g, /, etc., draw Hnes
parallel to BD. The distances A/',
Ak\ A^, etc., will then give the
original measurements to the en
^ Fig. 189. larged scale of 3 : 2.
To reduce the original drawing
in the same proportion, i.e. 2 : 3. With A as centre, and
radius AB, draw the arc BE. From E draw the line EF
parallel to BD. AF will then represent AB reduced in the
proportion of 2 : 3, and so on with any other measurement that
we may require.
EXERCISES.
1. Construct a scale to measure feet and inches; the R.F. to be tjVj
and its scale length value 15 feet.
2. Construct a scale to measure yards and feet, the R.F. to be ^V,
to measure 18 yards.
3. Construct a diagonal scale to measure feet and inches, R.F. yV,
to measure 36 feet. Take off a length of 17' 9".
4. On a map, a distance known to be 20 miles measures 10" ;
construct a diagonal scale to measure miles and furlongs, long enough
to measure 12 miles.
5. Construct a diagonal scale to measure yards and feet, R. F. ■^\^,
to measure 30 yards.
6. On a map showing a scale of kilometres, 60 are found to equal
3". What is the R.F.? Construct a comparative scale of English
miles, to measure 100 miles.
PLAIN, COMPARATIVE, AND DIAGONAL SCALES. 91
7. A line 4f " long represents a distance of 4'. Construct a scale by
which feet and inches may be measured up to 4 feet. The scale must
be neatly finished and correctly figured. (April, '96.)
8. Construct a scale to show yards and feet, on which 3^" represent
8 yards. Make the scale long enough to measure 10 yards, and finish
and figure it properly. (May, '97.)
9. Construct a scale of 7^' to i", by which single feet may be
measured up to 30'. The scale must be neatly finished and correctly
figured. (June, '97.)
10. The diagram represents an incomplete scale of feet (Fig. 190.)
Complete the scale so that distances of 2' may be
measured by it up to 50'. The scale must be ' ^ J^
properly finished and figured. (April, '98. ) Fig. 190.
11. Construct a scale onetenth of full size, to measure feet and
inches up to 5 feet. The scale must be properly finished and figured.
(June, '98.)
12. Construct a scale of feet and inches oneninth ( ^) of full size long
enough to measure 4 ft. The scale must be properly finished and
figured, and should not be "fully divided" throughout i.e. only one
distance representing i ft. should be divided to show inches. (April, '99. )
13. T he line AB (2.75 inches long) represents a distance of 2^ ft.
Make a scale by which feet and tenths of a foot may be measured up
to 4 ft. The scale must be properly finished and figured, and should
not be "fully divided " throughout, ?'.<?. only one distance representing
I foot should be divided to show tenths. (June, '99.)
14. The given line AB (3.2 inches long) represents a distance of 35 ft.
Construct a scale by which single feet may be measured up to 40 ft.
The scale is not to be "fully divided" i.e. single feet are not to be
shown throughout the whole length, and it must be properly finished
and figured, (April, '00. )
15. A drawing is made to a scale of i^" to i' , and another drawing
is required on which the dimensions shall be threequarters of those on
the fir.t drawing. Make a scale for the second drawing to show feet
and inches up to 5', The scale is not to be "fully divided" [i.e. only
one length of i' is to be divided to show inches) and it must be properly
finished and figured. (June, '00.)
16. Construct a scale \ of full size, by which feet and inches may be
measured up to 2 feet. Show also distances of ^" by the diagonal
method. Finish and figure the scale properly, and show by two small
marks on it how you would take off a distance of i' 7V'. (April, '01.)
17. Draw a diagonal scale, onetenth of full size, by which centi
metres may be measured up to one metre. Mark on the scale a length
of 47 centimetres.
One metre (100 centimetres) may be taken as 39". (^lay, '03,)
CHAPTER IX.
INSTRUMENTS FOR MEASURING ANGLES, ETC.
A protractor is an instrument used for measuring or setting off
angles ; it may be either semicircular or rectangular in shape, as
shown in Fig. 191. The point C marks the centre from which
the radiating lines are drawn, and corresponds with the centre
of the circle.
The degrees are numbered in primary divisions, equal to
ten degrees each, on the outside line from A; and on the inside
line from B. In the actual instrument each of these primary
divisions is subdivided into ten secondary divisions, each of
which represents one degree. Only one of these is divided in
the figure.
133. To construct a scale of chords.
A scale of chords is constructed in the following manner
(Fig. 192). Draw the Hues AC and CD perpendicular to each
INSTRUMENTS FOR MEASURING ANGLES, ETC. 93
other. With C as centre, draw any quadrant AED, and
divide the arc into degrees (only the primary divisions are
shown in the figure). Join AD. With A as centre, and each of
the primary divisions as radii, draw arcs cutting the chord AD,
which will form the scale
of chords. . ??
To use this scale in
setting off an angle — for
example, to draw a line
that will make an angle
of 40° with line CB (Fig.
192).
With C as centre, and
radius equal to A60 on
the scale of chords, draw an arc BFD. With a pair of dividers,
take the distance A40 from the scale, and set it off on the arc
BF from B. Join FC. Then FCB will be the angle of 40°
required.
Note. — A60 is always equal to the radius of the quadrant
from which the scale of chords is constructed.
134. To construct any angle without a protractor.
Draw CD perpendicular to AB. With C as centre, and
CA as radius, draw the semicircle ADB. Trisect the angle
DCB in E and F (Prob. 13). Trisect the angle ECB in H and
K (Prob. 14). Bisect FK
in L (Prob. 12). Then
DE = 3o°,EH = 2o°,HF=io°,
and FL = 5°. Therefore
between D and B we can
construct any angle that is a
multiple of 5°.
Divide the angle ACD into
five equal parts by the radii
from M, N, O, and P (Prob. 121). From A set off AR equal to
DE. As AN = 36° and AR = 3o°, ANAR = 6°, MR = 12°,
AM = 18°, and RO = 24°. Therefore between A and D we can
construct any angle that is a multiple of 6°,
tig. 193.
94 GEOMETRICAL DRAWING AND DESIGN.
If we subtract the multiples of 5° from those of 6° we can
obtain any desired angle, e.g. —
6 5 = 1°
1210 = 2°
1815 = 3°
2420 = 4°
3025 = 5°
etc. etc. etc.
All the regular polygons, with the exception of two — the
heptagon and undecagon — can be constructed with angles that
are multiples of 5° or 6°.
If the polygon is to be inscribed in a circle, the angle would
be set off at the centre of the circle ; but if one side of the
polygon is given, the angle would be set off externally, as
shown in Fig. 74, page 31.
The exterior angle of a Pentagon is 72° a multiple of 6°.
„ „ Hexagon „ 60° „ 6°.
„ „ an Octagon „ 45° „ 5°.
„ „ a Nonagon „ 40° „ 5°.
M i' Decagon „ 36° „ 6°.
„ „ Duodecagon „ 30° „ 5°and6°.
The Sector.
The Sector is an instrument of great utility for facihtating the
work of Practical Geometry. It consists of graduations on the
two radii of a footrule, and it is used by measuring the arc
between the graduations. Hence its name. The legs can be
opened to contain any angle up to a straight line.
In the illustration (Fig. 194) only the lines most used in
Practical Geometry are shown : viz. line of lines, marked L on
each leg ; a pair of lines of chords, marked C ; and a line of
polygons, m.arked POL, on the inner side of each leg.
The sectoral lines proceed in pairs from the centre of the
hinge along each leg, and although the scales consist of two
or three lines, parallel to the sectoral lines, all measurements
must be made on the inner lines of each scale, i.e. the lines
that radiate from the centre.
INSTRUMENTS FOR MEASURING ANGLES, ETC. 95
When the measurement is confined to a Hne on one leg of the
sector, it is called a lateral distance ; but when it is taken from a
Fig. 194.
point on a line on one leg to a similar point on a corresponding
line on the opposite leg, it is called a transverse distance.
Simple proportion. — Let AB and AC (Fig. 195) represent
a pair of sectoral lines, and BC and DE b,
two transverse measurements taken between
this pair of Hnes ; then AB is equal to AC,
and AD to AE, so that AB : AC : : AD : AE, d\   7£
and the lines AB : BC : : AD : DE.
The Line of Lines.
The primary divisions only are shown in
the illustration ; in the real instrument, each
of these is subdivided into ten secondarv divisions
135. To find the fourth proportional to three given lines.
From the centre, measure along one leg a lateral distance equal
to the first term ; then open the sector till the transverse distance
96 GEOMETRICAL DRAWING AND DESIGN.
between this point and a corresponding point on the other leg
is equal to the second term ; then measure from the centre
along one leg a lateral distance equal to the third term ; the
transverse distance from this point to a corresponding point on
the opposite leg will then give the fourth term.
Example. — To find the fourth proportional to the numbers 3,
4 and 9. From the division marked 3, which is the first term,
open the sector till the distance between this point and the
corresponding division on the other leg is equal to 4 divisions :
this will be the second term. Then 9 being the third term, the
transverse distance between the corresponding divisions at this
point will give the fourth term, viz. 12.
136. To find the third proportional to two ^ven lines or
numbers.
Make a third term equal to the second, then the fourth term
will give the required result
137. To bisect a given line.
Open the sector till the transverse distance between the end
divisions, 10 and 10, is equal to the given line ; then the transverse
distance between 5 and 5 will bisect the given Hne.
138. To divide a given line AB into any number of equal parts.
For example, eight (Fig. 196). When the number of parts is a
power of 2, the division
A ' D — ^ Q ' ' ' ^ is best performed by suc
Fig. 196. cessive bisections. Thus,
make AB a transverse
distance between 10 and 10, then the distance between 5 and
5 will give AC = half AB. Then make the transverse distance
between 10 and io = AC, the distance between 5 and 5 will
then give AD = one quarter of AB. By repeating the operation
each quarter will be bisected, and the given line divided into
eight equal parts as required.
INSTRUMENTS FOR MEASURING ANGLES, ETC. 97
When the number of divisions are unequal,— for example,
seven (Fig. 197), — make the transverse distance between 7 and
I 1 I I 1 I ' I
A c B
Fig. 197.
7 equal to the given line AB ; then take the distance between
6 and 6, which will give AC. The distance CB will then divide
the line into seven equal parts.
139. How to use the sector as a scale.
Example. — A scale of 1 inch equals 5 chains. Take one inch
on the dividers, and open the sector till this forms a transverse
distance between 5 and 5 on each line of lines ; then the corre
sponding distances between the other divisions and subdivisions
will represent the number of chains and links indicated by
these divisions : for instance, the distance between 4 and 4
represents 4 chains, 6.5 = 6 chains 50 links, Z7="3 chains 70
links, etc.
Note. — i chain is equal to 100 links.
140. To construct a scale of feet and inches, in which 2^ inches shall
represent 20 inches.
Make the transverse distance between 10 and 10 equal to 2\
inches ; then the distance between 6 and 6 will represent 12 inches.
Make AB (Fig. 198) equal to this length. Bisect this distance
' P . . ^ . t . , ^ . t ^ / ^.^
A D C E F B "
Fig. 198.
in C, as described for Fig. 196 ; then bisect AC and CB in D
and E in the same manner. Take the transverse distance
between 5 and 5, which will give AF 10 inches ; EF will then
trisect each of the four divisions already obtained. AB will then
be divided into twelve divisions, which will represent inches.
Produce the line AB to H, and make BH equal to AB. BH
will then represent one foot.
G
98 GEOMETRICAL DRAWING AND DESIGN.
141. How the sector may be used for enlarging or reducing a
drawing.
Let ABC (Fig. 199) represent three points in a drawing, let
it be required to reduce this in the proportion of 4 to 7.
Make the transverse distance between 7 and 7 equal to AB ;
then take the distance between 4 and 4, and make DE equal to
this length. Also make the distance between 7 and 7 equal to
A
Fig. 199.
AC ; then take the distance between 4 and 4, and from D as
centre, with this distance as radius, describe an arc. In the
same manner make the distance between 7 and 7 equal to BC ;
then with a radius equal to 4, 4, describe another arc from E,
cutting the other arc in F. Join EF and DF. Then DEF will
be a reduced copy of ABC, in the proportion of 4 : 7 as required.
142. To enlarge a drawing in the proportion of 7 to 4.
In this instance the sector would be opened so that the
transverse distance between 4 and 4 should represent the
original measurements, while those required for the copy would
be taken between 7 and 7.
The Line of Chords.
In the scale of chords already described (Prob. 133) we are
limited to one radius in setting off angles — viz. a radius equal to
the 60 marked on the scale ; in the double line of chords on the
INSTRUMENTS FOR MEASURING ANGLES, ETC. 99
sector there is no such hmitation — we can set off any radius
equal to the transverse distance between the two points 60 and
60, from their nearest approach to each other up to the fullest
extent the opening of the sector will admit of.
143. To construct an angle of 50^.
Open the sector at any convenient distance.
Take the
transverse distance between the
points 60 and 60, and construct an
arc with this radius. Let AB
(Fig. 200) represent this radius.
Now take the transverse distance
between 50 and 50, and set it off
from B on the arc, which will give
the point C. Join AC. Then BAC
will be 50°, as required.
A greater angle than 60° cannot
be taken from the sector with one measurement ; if the angle to
be measured is more than 60°, successive measurements must be
taken.
144. On an arc f inch in radius, to construct an angle of 125°.
Make the transverse distance between the points 60 and 60
I inch. Let AB (Fig. 201) represent this distance. Describe
an arc with AB as radius. Take
the distance between the points
50 and 50 from the sector, and
set it off on the arc from B to C.
Also take the distance from 40
to 40, and set it off from C to
D. Then take the distance
between 35 and 35, and set it
off from D to E. Join EA. Then the angle BAE will be 125°.
5o° + 4o° + 35°i25°.
145. To construct an angle of 3^ on the same arc.
With the sector open at the same angle as before, take the
transverse distance between the points 47 and 47, and set it off
on the arc from B to H. Join HA and CA. Then HAC will
be 3° as required. 50° 47° = 3°
loo GEOMETRICAL DRAWING AND DESIGN.
The Line of Polygons.
This pair of lines is used for dividing a circle into any number
of equal parts between four and twelve, by joining which the
regular polygons are formed. The transverse distance between
the points 6 and 6 is always used for the radius of the circle to
be divided ; because the radius of a circle containing a sixsided
figure, i.e. a hexagon, is always equal to one side of the figure.
Open the sector till the transverse distance between 6 and 6
is equal to the radius of the circle ; then the distance between
the points 4 and 4 will divide the circle into four equal parts,
the distance between 5 and 5 into five equal parts, and so on
up to twelve.
If it be required to construct a polygon on a given straight
line, open the sector till the transverse distance between the
numbers answering to the number of sides of the required
polygon shall equal the extent of the given line, then the distance
between the points 6 and 6 will give the radius of the circle to
be divided by the given line into the required number of equal
parts.
146. On a given line 1 inch in length, to construct a heptagon.
Open the sector till the transverse distance between the
points 7 and 7 shall equal i inch ; the distance between the
points 6 and 6 will then give the radius of a circle, to which
the given line will form seven equal chords.
EXERCISES.
1. On a line 4" long, draw a semicircle, and upon it set out the
primary divisions of a protractor, by construction alone.
2. Construct a scale of chords from the protractor set out in the
preceding question.
3. Make a scale of chords of 2" radius, to read to lO° up to 90°.
The scale must be finished and figured. At the ends of a line 2^" long
construct, from the scale, angles of 20° and 70° respectively. (April, '99. )
CHAPTER X.
THE CONSTRUCTION OF SIMILAR FIGURES.
PRINCIPLES OF SIMILITUDE.
Similar Figures.
Similar figures have their angles equal and their corresponding
sides proportional.
All regular figures — such as equilateral triangles, squares,
and regular polygons — are similar. Other quadrilateral figures —
triangles and irregular polygons — can be constructed similar to
given ones by making their angles equal.
147. To construct within a given triangle ABC, and equidistant
from the sides of it, a
similar triangle, the
base of which is equal
to the given line D,
Fig. 202.
Bisect the angles BAC
and ACB by lines meet
ing at the centre E (Prob.
12). Join EB. On the
line AB set off AF equal to
the given line D. From
F draw a line parallel to
AE till it cuts EB at G
(Prob. 3). From G draw
a line parallel to BC till it
cuts EC at H. From H
draw a line parallel to AC till it cuts EA at K.
KGH will be the similar triangle required.
Fig. 202.
Join KG.
102
GEOMETRICAL DRAWING AND DESIGN.
148. To construct about a given triangle ABC, and equidistant
from its sides, a similar triangle, the base of which is equal
to a given line L. Fig. 202.
Set off on the base AB produced, AN equal to the given line
L. From N draw a line parallel to EA (Prob. 3) till it meets
EB produced at O. From O draw a line parallel to AB till it
meets EA produced at M. From M draw a line parallel to AC
till it meets EC produced at P. Join PO. Then MOP will be
the similar triangle required.
To construct within a given square ABDC, and equidistant
150.
from its sides, a square, one
side of which is equal to the
given line E.
Draw the diagonals AD and
CB. From A set off AF along
AB equal to the given line E.
From F, parallel to AD, draw a
line till it meets CB at G (Prob.
3). With M as centre, and
radius MG, set off the points H,
K, and L, and join GH, HK,
KL,andLG. Then HGLKwill
be the square required.
Fig. 203.
To construct a triangle similar to a given triangle CDE, and
having its perimeter equal to a given straight line AB.
On the given line AB construct
a triangle ABF similar to the
given triangle CDE, by making
the angles at A and B equal to
the angles at C and D respec
tively. Bisect the angles at A
and B by lines meeting at G.
From G draw a line parallel to
FB till it meets AB at L ; and
also a line parallel to AF till it
meets AB at H. Then HLG
will be the triangle required.
CONSTRUCTIOX OF SIMILAR FIGURES.
103
Principles of Similitude.
Draw a rectangle ABDC, and join each angle to any
E. Bisect EB in G, EA in F, ED in K, and EC in H.
FG, GK, KH, and
HF, then FGKH
point
Join
will
be a rectangle with
sides one half the
length of the rectangle
ABCD. If we draw
the diagonals BC and
GH, we shall find
they are parallel to
each other, and that i"ig205.
BC : GH as 2:1. If we take any point L in BD and join
it to point E, LE will intersect GK in M, and will divide GK
in the same proportion as BD is divided.
On the principle of this problem, we can draw a figure similar
to a given figure, and having any proportion desired, e.g. —
If we wish to draw a rectangle having sides equal to one
third of a given rectangle, we should trisect the lines drawn
from the angles of the given rectangle to E. The point E is
called the centre of similitude of the two figures ABDC and
FGKH, which in this instance are said to be in direct similitude.
The following problem shows the principle of inverse similitude.
151. To draw a trapezium similar to a given trapezium ABDC,
with sides two thirds the length of those of the given
trapezium, In INVERSE SIMILITUDE.
Take any point E in
a convenient position.
Produce a line from A
through E to K, mak
ing EK equal to two
thirds of AE (Prob.
117). Proceed in the
same manner with each
of the points B, C, and
D, which will give the points H, G, and F. Join KH, HF, FG,
and GK. Then FGHK will be the trapezium required.
104
GEOMETRICAL DRAWING AND DESIGN.
152. To draw an irregular polygon similar to a given polygon,
but with sides twothirds the length of those of the given
polygon ABCDEFG.
It is not necessary that the centre of simihtude should be
taken outside the figure ; if more
convenient, we can use one of
the angles of the figure, e.g. let
A be the centre of similitude.
Draw lines from all the angles
of the polygon to A. Make AK
twothirds of AD (Prob. 117),
and divide all the other lines in
the same proportion, which will
give the points L, M, N, O, H,
andK. Join NM, ML, LK, KH,
and HO, which will give the polygon required.
Note. — The figures are similar, and the corresponding sides
Fig. 207.'
are parallel to each other.
153. To draw an irregular pentagon similar to a given pentagon
ABCDE, but with sides onehalf the length of those of
the given pentagon,
without using any
centre of similitude.
Draw FG in any
convenient position
parallel to AB, and
half its length (Prob.
4). From G draw
GH parallel to BC
and half its length.
Proceed in the same
way with the remaining sides, which will give the pentagon
required.
154.
To draw a curve or pattern similar to a given figure,
but to twothirds the scale.
Enclose the given figure in a convenient rectangular figure
ABDC, and divide the sides of the rectangle into equal parts
CONSTRUCTION OF SIMILAR FIGURES.
105
(Prob. 9). Join these divisions, which will divide the rectangle
into a number of equal squares or rectangles. Draw another
G
>
^
H
/
/
"^
,
/
^
,\
—
^
E
F
r
>
^
[
/
7
^
\
/
/
^
\
A
/
/"^
^:^V(^
^^
w
Fig. 209.
Fig. 210.
rectangle EFGH with sides twothirds the length of the
rectangle enclosing the given figure, and divide it in a similar
manner.
Draw the curves to intersect these smaller divisions in the
same places as the larger divisions are intersected by the given
figure.
Note. — This method is used for enlarging or reducing maps
or drawings to any scale.
EXERCISES.
1. Draw a regular hexagon on a side of i"; construct a similar figure
on a side of ^", using one angle as the centre of similitude.
2. Draw a rectangle with sides of 2 7" and 1.5" ; construct a similar
figure by inverse similitude, with sides in the proportion to those given
as 3 : 5
3. Draw a regular pentagon in a circle of 2§" diameter ; construct a
similar figure by direct similitude, with sides in the proportion to those
of first pentagon as 4 : 7.
4. Make an irregular pentagon ABODE from the following data :
Sides: AB=:2i", BC=ii", CD = 2r, DE=i".
Angles: ABC =105°, ABE = 30°, BAE=io5°.
Then make a si7Jiilar figure in which the side corresponding to AB is
If" long. (June, '98.)
io6
GEOMETRICAL DRAWING AND DESIGN.
5. Construct a rectangle hiavinq: one of ils sides if", and its diagonals
2" long. Make a similar rectangle having its shorter sides i^" long.
(April, '96.)
6. Construct a triangle, sides i^", i", and 2^" long, and a similar
triangle having its longest side 2^" long. Measure and write down the
number of degrees in each of the angles. (Tune, '97.)
7. Construct a figure (Fig. 211) similar to that given in the diagram,
but having the side corresponding to AB 2" long.
(May, '97.)
Fig. 212.
Fig. 213.
8. Construct a figure similar to the given figure (Fig. 212), but having
the distance corresponding to CD i" long. (June, '99.)
■ 9. Make a figure similar to the given one (Fig. 213), but having the
length corresponding to CD 2" long. (April, '00.)
CHAPTER XL
CONIC SECTIONS.
A conic section is obtained by intersecting a cone by a plane.
There are fiv^e different sections to a cone, viz. :
1. A triangle, when the plane cuts the cone
through its axis.
2. A circle, when the plane cuts the cone
parallel to its base, as at A, Fig. 214.
3. An ellipse, when the plane cuts the cone
obliquely, without intersecting the base, as at B.
4. A parabola, when the plane cuts the cone
parallel to one side, as at C.
5. An hjrperbola, when the cone is cut by a plane that is per
pendicular to its base, i.e. parallel to its axis, as at D, or inclined
to the axis at a less angle than the side of the cone.
These curves can be drawn with the greatest accuracy and
facility by the following arrangement. Cut a circular opening in
a piece of thin cardboard or stiff
paper, and place it a short dis
tance from a lighted candle ; this
will form a cone of light (Fig.
215). If we place a plane, e.g.
a piece of paper pinned to a
drawingboard, so as to allow
the hght coming through the
circular aperture to fall upon it,
we can, by placing it in the
several positions, intersect this cone of light so as to form
the required sections, which can then be traced. C is the
candle, A is the circular aperture, and P the plane.
GEOMETRICAL DRAWING AND DESIGN.
In Fig. 215 the plane is parallel to the aperture, so the section
obtained is a circle.
Fig. 216.
Fig. 217.
If the plane is placed obHquely to the aperture, as in Fig. 216,
the section obtained is an ellipse.
By placing the plane parallel to the side of the cone, as in
Fig. 217, we get as sec
^^''' tion, the parabola.
If we place the plane
at right angles to the
aperture, we obtain the
hyperbola, Fig. 218.
By adjusting the posi
tions of the candle, aper
ture, and plane, we can
required condition, both as
Fig. 2
obtain a conic section to suit any
to shape or size.
A truncated cone or frustum is the part of the cone below any
section as A or B, Fig. 214.
The Ellipse.
An ellipse has two unequal diameters or axes, which are
at right angles to each other. The longer one is called
the transverse diameter, and the shorter one the conjugate
diameter.
The transverse diameter is also called the major axis, as
AB (Fig. 219), and the conjugate diameter the minor axis,
as CD.
CONIC SECTIONS.
109
Fig. 219.
155. The two axes AB and
CD being given, to con
struct an ellipse.
Take a strip of paper and
set ofif upon it the distance
FH, equal to half the major
axis and the distance FG,
equal to half the minor axis.
By keeping the point G on
the major axis, and the
point H on the minor axis,
the point F will give a
point in the ellipse. A
succession of points can be
found in this manner,
through which draw a fair
curve, which will be the
required ellipse.
156. To construct an ellipse, given an axis and two foci.
An ellipse has two foci, as the points A and B, Fig, 220,
and the sum of the radii from these two points is always
equal.
Let A and B represent
two pins, and ABC a piece
of thread. The point of a
pencil is placed inside the
thread at C and moved
so as to keep the thread
always tight ; the point
will trace out an ellipse.
As the length of the thread
is constant, the sum of
the two radii is constant
also.
The length of the major
or minor axis given will
determine the length of the
thread.
Fig. 220.
no GEOMETRICAL DRAWING AND DESIGN.
157. To construct an ellipse by means of intersecting lines, the
transverse diameter AB and the conjugate diameter CD being
given.
Draw the lines AB and CD bisecting each other at right
angles in the point E (Prob. 5). Draw HK and FG parallel to
AB, through C and D,
H
W\
E ' ' /'
V^4
,2> /2. 'I
B
and HF and KG parallel
to CD through A and B.
(Prob. 3). Divide AH and
BK into any number of
equal parts, say four (Prob.
9), and AE and EB into
the same number. Join
C with the three points in
AH and BK, and produce
lines from D through the
three points in AE and
EB. Where these lines
intersect those drawn from C, points in onehalf of the ellipse
will be obtained. Find corresponding points for the other half
in the same manner, and draw a fair curve through the points
obtained, which will be the required ellipse.
D
Figr. 221.
158. To find the normal and tangent to a given ellipse ABCD, at a
given point P.
With C as centre, and
radius equal to EA, draw
the arc FH, which will
give the two foci in F and
H. Join the given point
P with F and H, and
bisect the angle FPH by
the line PK (Prob. 12).
PK is the normal. Draw
the line NO through P,
perpendicular to PK.
This is the tangent re
quired.
CONIC SECTIONS.
Ill
159. To complete an ellipse from an elliptical curve.
Let AB be the given curve. Draw any two sets of parallel
Join the points of bisection
v^F
chords and bisect them (Prob
in each set by lines meeting
in C. Produce one of these
lines till it meets the given
curve in D. With C as y^VA \ ^M
centre, and CD as radius, set
off on the given curve the
point A. Join AD. Through
C draw a line HK parallel
to AD (Prob. 3), also the ^
line CL perpendicular to AD
(Prob. 7). Produce CL to
M, making CM equal to CL.
Also make CK equal to CH.
Then LM will be the transverse and HK the conjugate
diameters. The ellipse can then be completed by any of the
constructions already described. From A and D lines are drawn
parallel to LM ; with C as centre, and radius CD, set off E
and F, these will be two more points in the ellipse.
160. To draw an ellipse to pass through three given points A, B, C.
Join AC and bisect it in D. Join BD. From A and C draw
the lines AE and CF parallel to BD. Through B draw the line
EF parallel to AC. E B
Produce BD to H,
and make D H equal
toBD. Divide AD
and DC and also
AE and CF into a
number of equal
parts, say four. Join
the divisions on AE
andCFtoB. From
H, through the divisions on AC, draw lines till they meet the
corresponding lines drawn to B. Draw a fair curve through
these points, which will give half of the required ellipse. Proceed
in the same way with the other half.
112 GEOMETRICAL DRAWING AND DESIGN.
161. To determine points for drawing a parabola, the focus A
and the directrix BC being given.
Draw the line EAD perpendicular to the directrix BC (Prob. y),
which will give the axis. Bisect AD in F (Prob. i), which will
be the vertex of the curve. Take any points a, d, c, d, and e in
the axis, and draw perpendiculars through them. From A as
centre, mark off on the perpendiculars, arcs with radii equal to
aD, dD, cD, dD, and eD, cutting the perpendiculars in a'^ b\ ^,
d\ and e' . These are the points required for the lower half of
the parabola. The points above the axis are found in the same
manner.
B
V
A
^
K/^
■^
f
/'
1
\
E
c
A
\
b
C
a'
d L
e
Fig. 225.
162. To draw a tangent to a parabola at a given point H.
Join AH. From A set off AK on the axis produced equal
to AH. Join KH, which will be the required tangent. This
could also be found by drawing a Hne from H parallel to the
axis till it meets the directrix in B, and then bisecting the angle
AHB by the line KH (Prob. 12), which is the tangent. If from
H we draw the line HL perpendicular to the tangent, it will be
the normal.
CONIC SECTIONS.
113
163. To draw a parabola, an abscissa AB and an ordinate BC
being given.
Complete the rectangle ABCD. Divide BC into any number
of equal parts, say six
(Prob. 9), and CD into
the same number. From
each division in BC draw
lines parallel to CD (Prob.
3), and from each of
the divisions in CD draw
lines to the vertex A.
Where these lines of
corresponding numbers
intersect, e.g. where i
intersects with i', 2 with
2', etc., are points in the
parabola. Find corre
sponding points on the opposite side of the axis, and draw a
fair curve through them.
164. To draw an hyperbola, the diameter AB, an ordinate CD,
and an abscissa BD being given.
Draw BE parallel to CD (Prob. 3), and complete the rectangle.
Produce BD, and make
AB equal to the given
diameter. Divide CD
and CE into any number
01 equal parts, say four
(Prob. 9), «, ^, c. The
divisions on CD join to
A, and those on CE to B. c'
The intersection of the u
corresponding lines, e.g. ,
where a intersects c^', b b\ ^
and c c\ are points in the
hyperbola required. Find
corresponding points for
the other half, and draw a fair curve through them.
H
114
GEOMETRICAL DRAWING AND DESIGN.
A form of hyperbola frequently used is the rectangular hyper
bola. Let AB and AC
represent two axes, and
E the vertex of the curve.
Complete the rectangle
ABDC. Take any point
H in CD and join it to
A. Let fall a perpendi
cular from E till it meets
HA in O (Prob. 7).
From O draw OK parallel
to AB till it meets a line
from H parallel to AC in the point K (Prob. 3). This will be
one point in the curve, and others may be found by taking fresh
points on CD and treating them in a similar manner.
One peculiar property of this figure is that, if we take any
point in the curve and draw lines from it perpendicular to the
lines AB and AC, — for example, KN and KM, — the rectangle
contained by the two lines is always equal, i.e. KN x KM would
be the same for any point in the curve.
165. A mechanical method of drawing a parabola or hyperbola.
Let AB represent the edge of a drawingboard and CD the
edge of a teesquare. Take
a piece of string equal in
length to CD, fix one end at
D and the other at E, which
is the focus of the curve. If
a pencil be held against the
string so as to keep it tight
against the tee square when
the teesquare is moved up
wards, the pencil will trace
half a parabola. AB is the
directrix, and K the vertex of
Fig. 229. the curve. Compare this
method with the construction of Prob. 161.
If the angle DCA were an acute or obtuse angle instead of a
right angle, the pencil would trace an hyperbola.
CONIC SECTIONS.
"5
166. To draw an oval by arcs of circles, its transverse diameter AB
and its conjugate CD being given.
Set off on AB the distance AE equal to half the conjugate
diameter. Through E draw the line FG perpendicular to
A
AB (Prob. 7). With E
as centre and EA as
radius, draw the semi
circle CAD. From C
and D set off CF and
DG equal to EA. From
B set off BH equal to
half of EA. Join FH
and GH. With F and
G as centres, and FD,
GC as radius, draw the
arcs DL and CK. With
H as centre, and H K as
radius, draw the arc
/
c/
^
E \
V i
1
1
>
/ '
\N
X /
\ ^
^ /
H^' /
\ X
\ ^
N X
*K^
j/l
D C,
B
Fig. 230.
KBL, which wili complete the oval required.
The conic sections are of frequent occurrence both in science
and art : the heaAcnly bodies trace them in their courses ; they
are used by engineers where A ^
great strength is required, such
as the construction of bridges ;
and they form the contour of
mouldings, etc. Those subtle w
curves that we admire in the
outline of Japanesehandscreens
and vases are often parabolas.
Fig. 231 is an illustration
showing how these curves are
applied to art forms.
To draw the curve CB.
Draw the lines AB and AC
at right angles to each other
(Probi* 5). Di^id^^ch into
the same number of equal jjarl#^.^,
(Prob. 9), and join them. Pro
Fig. 231,
ceed in the same manner with the other curve**
ii6 GEOMETRICAL DRAWING AND DESIGN.
Cycloidal Curves.
If a circle is rolled along a line in the same plane, a point in
the circle will describe a curve of a class called cycloidal.
The line along which the circle rolls is called a director, and
the point itself is called the generator.
The curve is called a Cycloid when the generator point is in
the circumference of the rolling circle and the director is a
straight line ; but a Trochoid when the point is not in the circum
ference of the circle.
When the director is not a straight line, but the outside of
another circle, and the generator is in the circumference of the
rolling circle, the curve desc^bed is called an Epicycloid ; but
when the point, or generator, is not in the circumference of the
rolling circle, it is called an Epitrochoid.
If the director is the inside of a circle and the generator a
point in the circumference of the rolling circle, the curve is
called a Hypocycloid ; but if the generator is not in the circum
ference of the rolling circle, it is called a Hypotrochoid.
The Epicycloid and Hypocycloid are the true curves for the teeth
of gearing. The director is the pitch circle of each wheel : and
if the rolling circle be the same for the whole set, they will gear
into one another.
In constructing a cycloid it is necessary to make a line equal
in length to the arc of a semicircle. The exact relation betwee"
the diameter and circumference of a circle cannot be express
in numbers ; but the following problem will enable us to arri
at an approximation, correct to six places of decimals.
167. To draw a line equal to tlie length of a semicircle.
Let AC represent the radius. Draw the semicircle ABD.
Produce AC to D, and draw BC perpendicular to it. From A
and D draw tangents parallel to BC, and through B drav/ a
tangent parallel to AD. From B set off ;S£>»e(ifiial to the jadius,
and draw the line BF through E« Pjoduce the tangent through
A to H, and makej^H equal to AD. Join HF, which will be
the line require^
CONIC SECTIONS.
If we take AC to represent a length equal to the
of a circle, then HF will equal the circumference,
to the dimneter
F D
Fig. 232.
We are also enabled to find the length of an arc by this
means ; e.g: the arc to the chord formed by one side of a pentagon,
' irfH IS eq3.l.tn j^he circumference, then
arc required.
FH
thelenofh(;,;\,,e
168. To draw a cycloid.
Let AB be the director, and / the generator or point in the
rolling circle AM/). Draw Ak equal in length to half the
circumference of the circle AM/, and divide it into any number
^ "" equal divisions (Prob. 10), say six, at d, e,/, j^, and /t. Divide
the semicircle into the same number of equal divisions (Prob.
14), and draw lines from each division parallel to the director
AB. Draw the line CK from the centre of the circle parallel to
AB. Draw lines perpendicular to AB at the points d, e,/,g; /i,
and k till they meet the line CK in the points D, E, F, G, H,
and K. With each of the points D, E, F, G, H, and K as
mtres, and a radius cqunl to Cp, draw arcs cutting the parallel
^^»f^l^m^tltei the dnisions ii^'tb^ fliiftlicircle in the points
K^ ^f f^,f3lV »hid T. This wilT give points in half the cycloid.
ndf:<^ 40rt^s(pWiciwig' rtcjfnts for the remaining half, and draw
feU'CtiWfe' through tlve p«hir ;v^^liH give the cycloid
furtTe'd.
ii8
GEOMETRICAL DRAWING AND DESIGN.
To determine the tangent and normal to the curve at any
point /:— Draw the Hne //' parallel to AB till it meets the
Fig. 233.
generating circle at /'. Join /'A. Through / draw the line
WW parallel to /'A. This will be the normal to the curve. The
tangent rs is at right angles to this line.
169. To draw an epicycloid.
Note. — The length of the di
a complete curve is
to the whole circle as the radius of the rolling circle is to the
radius of the director; e.g. if radius of rolling circle == i inch, and
that of director = 6 inches ; then the director = ^ of the circle.
Let AB be the director, which is a part of a circle, and/ the
generator. Take Ak equal in length to half the rolling circle,
AM/, and divide it into any number of equal divisions, say six,
at ^, ^, /, g., and //. Divide the semicircle into the same number ;
of equal divisions, and draw lines from these points, as well as
from the centre of the circle, concentric with the arc /:.B.
From the centre of the circle that contains the arc AB draw
lines through the points d, e,f,g., h, and k till they meet the arc
drawn from the centre of the rolling circle. With D, E, F, G,
H, and K as centres, and a radius equal to Cp, draw arcs til)
they meet the concentric arcs drawn from the divisions of th'^e
semicircle in the
sponding points^
through all the
At any poi^i^f ^^WF"""^BHHm^^^^^K c .^^^'
proceed ^,gJpllow^. ^R'^ ^^^ arc tt concent^^MPp^'AB»
t:> iy(X
CONIC SECTIONS.
119
till it meets the generating circle in /'. Join /'A. With / as
centre, and radius equal to /'A, draw an arc intersecting AB at
Fig. 234.
w'. Join tu>\ and produce it to w.
tangent rs is at right angles to it.
This is the normal. The
170. To draw a hypocycloid.
Let AB be the director, which is the arc of a circle, and p
the generator, which is a point in the circumference of the
roUing circle M. Make A/j equal in length to half of the circle
M, and divide it into any number of equal parts, say six, at d^
^■> /■> .^1 arid h. Divide the semicircle into the same number of
equal parts, and from, the centre of the circle containing the arc
AB draw concentric arcs from these points, ^s well as from the
centr
. e C. Draw lines from the points d, e^f, g, h^ and k towards
the centre of the circle co^maining the arc AB till they meet the
\rQ from the centre C in the points D, E, F, G, H, and K. With
 jt oe^ noints as centred, and a radius equal to C/, draw
res till they meet the concentric arcs drawn from the divisions
■ the semicircle in the points N, O, p," O, R, and T. Find
: corresponding points for the othef half, and draw a fair
•/e through all the points, which will give the hypocycloid
juired.
I20 GEOMETRICAL DRAWING AND DESIGN.
The tangent and normal at any point / are thus o i. ed.
Draw the arc W concentric to the arc AB till it meets ihe
generating circle at /'. Join /'A. With / as centre, and radius
Fig. 235.
equal to /'A, set off on AB the point w'. Join fw', and produce
it to 7U. This is the normal. The tangent rs is at right
anoles to it.
171. To construct a continuous curve, by a combination of arcs of
different radii, through a number of given points A, B, C, D, E,
F, G, and H.
Join the points AC, CD, DE, etc. Find the centre K of the
circle containing the arc
ABC (Prob. 35). Join CK.
Bisect the line CD at right
angles, and produce the
bisecting perpendicular till it
meets CK produced in L.
Join DL. Bisect the line''^]g
'^i^'A produce The oisecting
\ierpendicular till it meets
DL produced in '^l. Find
the remaining points N, O,
and P in the same manner.
'^' '^ ■ • The points K, L, M, N, O,
and P are the centresr^of the circles containing the arcs
necessary for joining the given points. ;
CONIC SECTIONS.
njiijdf^Q construct an Arcliimedean spiral of one revolution.
Draw a circle and divide it by radii into any number of equal
parts, say twelve (Prob. 9) a^ b^
c, d^ etc. Divide the radius o
into a corresponding number
of equal parts i, 2, 3, 4, etc.
(Prob. 9). From the centre of
the circle, with radius i, draw
an arc till it meets the radius a
in A, and from 2 till it meets
the radius b in B, and so on till
the whole twelve are completed.
Draw a fair curve through these
points, A, B, C, D, etc., which
will give the spiral required and
proceed in exactly the same manner for further revolutions.
173. To draw the logarithmic spiral..
The logarithmic spiral was discovered by Descartes,
also called the equiangular
spiral, because the angle the
curve makes with the radius
vector is constant. The curve
also bears a constant proportion
to the length of the radius
vector
Take any line AC for the
radius vector, and bisect it in D
(Fig. 238). With D as centre,
and radius DA, draw the semi
circle ABC. From the points
A and C draw any two lines AB
and CB cutting the semicircle.
Then ABC is a rightangled
triangle.
Bisect the Hne BC in E.
With E as centre, and EB as
"''^ius, draw the semicircle
It is
GEOMETRICAL DRAWING AND DESIGN.
BFC. Make the angle BCF equal to the angle ACB, and pro
duce the line till it meets the semicircle in F. Join BF. The
triangle BFC is then similar to the triangle ABC. By repeating
this construction we obtain a succession of similar triangles
radiating from a common centre C, and all forming equal
angles at this point. The exterior points of these triangles, viz.
A, B, F, M, N, O, etc., are points in the required spiral.
As each triangle with its curve forms a similar figure, it is
evident that the curve must form a constant angle with its
radius vector,' i.e. the line radiating from C, and the portion of
the curve accompanying each triangle, must also bear a constant
proportion to the length of its radms vector.
If we bisect the angle ABF by the dotted line HB, this line
will be the normal to the curve ; and the line KL, being drawn
at right angles to HB, is the tangent to the spiral.
As all the angles at C are equal, the spiral could be con
structed with greater facility by
first drawing a circle and divid
ing it into an equal number of
parts by radii, as shown in
Fig. 239.
Let AC be the radius vector.
Bisect it in D. With D as
centre, and radius DA, draw an
arc cutting the next radius CB
in B. Proceed in the same
manner with each radius in suc
cession, which will determine
the points H, K, L, M, N, O,
P, etc. Draw a fair curve
through these points, and we shall obtain a logarithmic spiral.
The greater the number of radii used in the construction,
the larger will be the angle BAC ; but the angle ABC will
always be a right angle, as will be seen by the construction in
Fig. 238.
Fig. 239.
174. To draw a spiral adapted for the Ionic volute by means of arcs.
Divide the given height AB into eight equal parts (Prob. 9).
Bisect the fourth part in the point C (Prob. i), and from it
CONIC SECTIONS.
[23
draw a line perpendicular to AB (Prob. 7).
equal in length to four of the divisions of AB
the eye of the volute D. This
is shown to a larger scale at E.
With D as centre, draw a circle
with a radius equal to C4. In
scribe a square in this circle
(Prob. 56), and bisect each of
its sides in the points i, 2, 3,
and 4 (Prob. i). Join these
points, and draw diagonals.
Divide each semidiagonal into
three equal parts and join them
(Prob. 9), thus making three
complete squares parallel to
each other. The corner of
each of these squares in succes
sion will be the centre of one
of the arcs, commencing at i, with
dotted hnes and arrow heads.
Make this line
which v/ill sive
L as radius, as shown by
EXERCISES.
1. Construct an ellipse: major axis 3.75", minor axis 2.25". Select
any point in the curve, and draw a tangent to it.
2. Construct an ellipse ; the foci to be 2^" apart, and the transverse
diameter 3f".
3. Draw a rectangle 3.25" x 2.3", and inscribe an ellipse within it.
4. Draw a parallelogram 3I" x 2", two of its angles to be 60° ;
inscribe an ellipse within it.
5. Construct an ellipse by means of a paper trammel ; the transverse
diameter being 4I", and the conjugate diameter 3" (Prob. 155).
6. Draw the two diameters of an ellipse each 3" long, and at an
angle of 45° with each other ; complete the ellipse.
7. Make a tracing of the ellipse given in question 3, and find the
diameters, foci, tangent, and normal.
8. With an abscissa 3" long and an ordinate 2" long, construct a
parabola.
9. With a diameter 1.4", an ordinate 1.8", and an abscissa 1.4",
construct a hyperbola.
124 GEOMETRICAL DRAWING AND DESIGN.
10. Draw a rec'angle 3" x 2", and let two adjacent sides represent
the axes of a rectangular hyperbola ; measure off along one of its longer
edges ^", and let this point represent the vertex of the curve ; complete
the hyperbola.
11. Draw a line 4" long to represent an abscissa of a parabola ; at one
end draw a line 3" long, at right angles to it, to represent the directrix ;
from the directrix, along the abscissa, set off i" to mark the focus ;
complete the parabola. At any point in the curve, draw a tangent and
normal to it.
12. Draw geometrically an ellipse, a parabola, a hyperbola, each
2 inches long, and a cone 2 inches high, and write the name to each.
Show the following five sections on the cone — a horizontal section, a
vertical one, not through the apex, and one through the apex, one
parallel to one side, and one cutting both inclined sides. Name figure
each section makes. (May, '96.)
13. The foci of an ellipse are 2" apart and its major axis is 3^" long.
Describe half the curve. (April, '96.)
14. The foci of an ellipse are 2" apart, and its minor axis is 2" long.
Draw the curve, and draw also a tangent from a point on the curve \"
from one of the foci. (June, '00.)
15. Draw the curve (Fig. 241) every point of which is at equal
distances from the line PQ and the point F. The curve, which is a
parabola, need not be shown below the line ST.
(April, '96.)
Dimensions to be trebled. Fig. 242.
Fig. 241.
16. Draw a straight line AB 3" long. Bisect AB in F. At F draw
FV f" long at right angles to AB. F is the focus, and V the vertex of
a parabola, A and B being points on the curve. Draw the curve from
A to B, showing the construction for at least 4 points. (June, '00. )
17. Two conjugate diameters of an ellipse are 3^" and 2" long
respectively, and cross one another at an angle of 60°. Draw the curve.
(April, '99.)
18. An arch in the form of a semiellipse is 6' wide and 2' high.
Describe the curve, and draw two lines perpendicular to it from two
points on the curve, each 2' from the top point of the arch.
Scale (which need not be drawn) ^" to i'. (April, '98.)
CONIC SECTIONS. 125
19. Draw the arch opening shown (Fig. 242), using the figured dimen
sions. The curve is a semielUpse, of which I'Q is a diameter, and RS
is half its conjugate character. (June, '99.)
20. Describe the spiral of Archimedes of three revolutions, whose
radius is 2 inches. (April, '98.)
21. Within a circle of 2" radius describe a ' spiral of Archimedes ' of
one revolution. (June, '99.)
22. Sketch the three sorts of spiral, and explain how each is gene
rated, and illustrate each sort by shells, plants, or animals. (May, '96.)
PART IL
SOLID GEOMETRY.
CHAPTER XII.
INTRODUCTION.
In the preceding subject, Plane Geometry, we have been
restricted to figures having length and breadth only, but Solid
Geometry treats of figures that have thickness in addition to
length and breadth.
The objects taken to illustrate the principle of this subject
are described under the head of Definitions, Solids (page lo).
By means of Practical Solid Geometry we are enabled to
represent on a plane — such as a sheet of paper — solid objects in
various positions, with their relative proportions, to a given scale.
Let us take some
familiar object, a
dressingcase for in
stance, ABDC, Fig.
243, and having pro
cured a stiff piece
of drawingpaper
HKLM, fold it in a
line at X, parallel to
one of its edges ;
then open it at a right angle, so that HX will represent the
edge of a vertical plane, and XL the edge of a horizontal plane ;
the line at X, where the two planes intersect, is called the line
of intersection, intersecting line, or ground line ; it shows where
INTRODUCTION.
127
the two planes intersect each other, and is generally expressed
by the letters X and Y, one at each end.
Having placed the dressingcase on the horizontal plane,
with its back parallel to the vertical plane, let us take a pencil
and trace its position on the jj g
horizontal plane by drawing a
line along its lower edges ; also
its shape on the vertical plane.
This can be done by placing
the eye directly opposite each
of its front corners in succes
sion and marking their apparent
position on the vertical plane,
and joining them. Having done
this, we will remove the dress
ingcase and spread the paper
out flat upon a table : this is
shown in Fig. 244. We have now two distinct views of the
object. The lower one is called a PLAN, and represents the
space covered by the object on the horizontal plane, or a view
of the dressingcase seen from above. The upper view shows
the space covered on the vertical plane, and is called an
ELEVATION : it represents the front view of the object.
In Solid Geometry all objects are represented as they would
appear traced or projected on these two planes at right angles
to each other : they are b
called coordinate planes.
/
J'
B
X
c
d
a
b
L
M
Fig. 244.
It is not necessary that
the object should be
parallel to them, as in
Fig. 243 : we can arrange
it in any position, making
any possible angle with X
either plane, but the hne
connecting the point on
the object with its respec
tive plane must always be perpendicular to that plane. We
shall understand this better if we refer to Fig. 245, in which we
will imagine the dressingcase suspended in midair, with its back
128
GEOMETRICAL DRAWING AND DESIGN.
still parallel to the vertical plane, but its under side inclined
to the horizontal plane. We will now trace it on each plane as
before, then by spreading the paper out flat we get a drawing
as shown in Fig. 246.
The student should compare Fig. 243 with Fig. 244, as well
as Fig. 245 with Fig. 246, so as to thoroughly understand
the relation between the coordinate planes.
The lines Aa\ B^', Y.e (Fig. 246), are all perpendicular to the
vertical plane ; and the lines A«, B/^, Cc, etc., are perpendicular
to the horizontal plane. These
lines are called projectors, and
are here represented by dotted
lines. The points in which
these lines meet the two co
ordinate planes are called pro
jections : if they are on the
vertical plane they are called
vertical projections, and if on the
horizontal plane horizontal pro
jections, of the different points ;
e.g. a! i§ the vertical projection of
point A, and a is its horizontal
projection. The length of the
horizontal projector shows the
distance of the point from the vertical plane, and the length
of the vertical projector its distance from the horizontal plane.
This method of representing solid objects by projection on
two planes is called orthographic projection, and is described
more in detail in Chapter XIV. As the projectors are parallel
to each other, it may also be called parallel projection.
All through this subject the points of the object are dis
tinguished by capital letters, as A, B, etc., while their horizontal
projections are represented as <2, b., etc., and their vertical pro
jections as «', b\ etc. ; by this means we are enabled to dis
tinguish the plan from the elevation. v.P. will also be used to
express the vertical plane, and h.p. the horizontal plane ; the
letters XY will always stand for the ground line.
The student should take particular notice that the lower points
in the plan always represent the front points in the elevation.
Fig. 246.
INTRODUCTION
129
It is not necessary to have an object to trace ; if we know
its dimensions, and its distances from the two planes, we can
construct the plan and elevation as shown in Figs. 244 and 246.
Figs. 243 and 245 are perspective views, and Figs. 244 and 246
are geometrical drawings of the same object. If the latter
were drawn to scale, we could find out the length, breadth, and
thickness of the object from these drawings.
Each perspective view is supposed to l3e taken from one
fixed point, i.e. the eye ; and lines drawn from different parts of
the object converge towards the eye considered as a point.
These lines represent rays of light from the object, and are
called visual rays : they form a cone, the vertex of which is the
position of the eye ; consequently, Perspective is called conical,
radial, or natural projection, because it represents objects as they
appear in nature. It is impossible to see an object as it is
represented by orthographic projection.
We will now take four simple solids, viz. a cube, a rectangular
solid, a pyramid, and a triangular prism, and show the different
positions they can occupy with reference to the coordinate
planes, i.e. the v.p. and H.P.
Fig. 247 represents the four solids in what is called simple
positions, i.e. parallel to both the v.p. and H.P.
H
'nx
D
Fig. 247.
A is the plan of the cube and A' its elevation.
B „ rectangular solid „ B' „
C „ pyramid „ C „
D ., triangular prism „ D' „
130 GEOMETRICAL DRAWING AND DESIGN.
Fig. 248.
Fig. 248 represents the same solids with their bases on the H.P.
as before, but their sides are now inclined to the V.P.
Fig. 249.
Fig. 249 shows them with their fronts and backs, parallel to the
V.P., as in Fig. 247, but with their bases inclined to the H.P.
Fig. 250.
INTRODUCTION.
Fig. 250. — They are here represented inclined to both the v.p.
and H.P., but they still have one set of edges parallel to the H.P.
Fig. 251. — Here they are shown with every line inclined to both
planes : instead of having one ecige resting on the H.P., as in
Fig. 25
Fig. 250, they are each poised on a corner. To distinguish this
position from the one illustrated in Fig. 250, we will call it com
pound oblique; although Figs. 250 and 251 generally come under
one head, as objects inclined to both planes^
CHAPTER XIII.
SIMPLE SOLIDS IN GIVEN POSITIONS TO SCALE.
Note, — Feet are indicated by one dash, and inche5i by two
dashes : thus — 3' 6" represents 3 feet 6 inches.
The student should draw the problems in Solid Geometry
to a scale three times that of these figures.
175. To project a quadrilateral prism 5" x 2^" x 2h" with one of its
smaller faces on the H.P., parallel to the v.p., and " from it.
Scale } full size. Fig. 252 a.
First draw the line XY ; then draw the plan abc'c;^  inch
below it. Draw perpendicular lines above XY, immediately
over the points a and 6,
^l>' 5" in height, which give
the points a' and b'. Join
a'b'. This is the elevation
of the solid.
176. To project the same
solid with one of its
longer faces resting on
the H.P., parallel to
the V.P., and l" from
it, to the same scale.
Fig. 252 B.
Draw the plan e/^^^,
S"X2^", and if inches
below XY. Draw perpendiculars above XY, 2^" high, and
directly over the points e and /, which will give the points ^'
and/'. Join <?' and/, which completes the elevation.
Fig. 252.
SIMPLE SOLIDS.
^33
The student should now project the four sohds illustrated
in Fig. 247 in the positions there shown, but to the following
dimensions and scale :
A to have a base 4" x 4", to be 4" high and 2h" from v. p.
B „ „ 8"x4" „ 2" „ 2^" „ v.P.
C „ „ 4"X4" „ 8" „ 2i" „ v.P.
D
6"X4" „ 4"
Scale J full size.
v.P
177. To project a quadrilateral prism 10 " x 5" x 5", with one of its
smaller faces on the h.p., at an angle of 45' with v.p., and
one edge 3^" from v.P. Scale ^^ full size. Fig. 253 A,
DrawXY. Take
the point ^, 3^"
below XY, and
draw the square
abed below this
point with its sides
at an angle of 45°
with XY. This
will be the plan.
Erect perpendicu
lars 10" high above
XY, and directly
over the points a,
b, and c. Join the
tops of these per
pendiculars, which completes the elevation.
I
'
g
' t
r
X
Y
A
',d
a
c g
r"
y
f
/
I 253
h
',
178. To project the same solid lying on one of its longer faces on
the H.P. with its longer edges forming an angle of 30° with
the V.P., and one of its corners l" from v.P., to the same
scale. Fig. 253 P..
Draw the point /i, i" below XY, and construct the plan £/(^/i
at the required angle below this point. Erect perpendiculars
5" high above XY, immediately over the points o^ e, and /.
Join the tops of these perpendiculars, which completes the
elevation.
134
GEOMETRICAL DRAWING AND DESIGN.
The student should now project the four soHds illustrated in
Fig. 248, in the positions there shown, but to the following
dimensions and scale :
A to have a base 5" x 5", to be 5" high, with one side inclined at
an angle of 30° with the V.P., and 2" from it.
B to have a base 10" x 5", to be 25" high, with both sides inclined
at an angle of 45° with the V.P., and 2" from it.
C to have a base 5" x 5", to be 10" high, with one edge of the base
making an angle of 60° with the V.P., and its nearest point
2" from it.
D to have a base 8" x 5", to be 5" high, with both sides of its base
making an angle of 45° with the v.?., and the nearest point
2" from it.
179. To project a quadrilateral prism 7i"x3"x3", resting on
one of its shorter edges on the H.P., and with its longer edges
parallel to the v. P., but inclined at an angle of 60° to the H.P. ;
one of its faces to be 2^" from v.p. Scale J full size. Fig. 254 B.
Draw XY. At point e' draw the elevation e'a'h'c' at the
required angle. 2J" below XY draw the line df parallel to it.
SIMPLE SOLIDS.
m
Let fall lines from a', b', and c\ at right angles to XY, and
make da and /c each 3" long. Join ac^ which completes the
plan.
The student should now project the four solids illustrated in
Fig. 249 from the following conditions :
A to be 2^" X 2^^" X 2^", with its base inclined at an angle of 45°
to H.P. ; to be parallel to the v. p., and 2^" from it.
B to be 5"x2V'x i^", with its base inclined at an angle of 30°
to H.P. ; to be parallel to the v.p., and 2^" from it.
C to be 2^"x2^"x 5", with its base incHned at an angle of 45°
to H.P. ; to have the edge of its base parallel to the v.p.,
and 2h" from it.
D to be 4" X 2" X 2", with its base inclined at an angle of 30°
to H.P. ; to have its ends parallel to v.p., and if" from it.
Scale ^ full size.
Before proceeding with the next problem. Fig. 254 B, it will be
necessary to understand thoroughly the angles which the solid
forms with the coordinate planes.
The longer edges are still inclined to the H.P. at an angle of
60° ; but instead of being parallel to the v.p., as in Fig. 254 A, they
axe in planes inclined to the
V.P. at an angle of 45°. This
does not mean that they
form an angle of 45° with
the V.P. Let us illustrate this
with a model.
Take a sheet of notepaper
and draw a diagonal to each
of its inside pages, as ad and
dc, Fig. 255. Now stand it
on a table against the 45°
angle of a setsquare. The
two pages will then represent
two planes at an angle of 45°
with each other. Let the
page a represent the V.P.,
and the line 6c one of the edges of the solid. The angle which
dc makes with the page a will be considerably less than 45°.
Fig. 25s
136 GEOMETRICAL DRAWING AND DESIGN.
180. To project a quadrilateral prism 7h" x 3" x 3", resting on
one of its shorter edges on the H.P., with its longer edges
inclined to the h.p. at an angle of 60°, and in vertical planes
inclined to the v. p. at an angle of 45° ; one of its lower comers
to be 1;^" from v.p. Scale  full size. Fig. 254 B.
Find the position of point n, i^" below XY, and draw the
lines nm and nk at an angle of 45° to XY. Make nm and n/c
equal in length to ^ and da (Fig. 254 a). Draw the line k/
parallel to 7?m, and of the same length. Join /;;z. Make k^
and /o equal to ae and cd (Fig. 254 a). Draw^// and o^ perpen
dicular to /^/. This will complete the plan. As every point in
the elevation is found in precisely the same way, it is only
necessary to explain the projection of one point : <?', for example.
Draw a perpendicular from o on plan till it meets a hori
zontal line drawn from d' (Fig. 254 a). This gives the position
of point o'.
The student should now project the four solids shown in Fig.
250 from the following conditions :
A to be 4"x4"x4", with one set of edges parallel to H.P. ; its
other edges to be inclined to h.p. at an angle of 45°, and
in vertical planes inclined to v.p at an angle of 30° ; its
nearest corner to be 4" from v.p.
B to be 8" X 4" X 2", with one set of edges parallel to H.P, ; its
longest edges inclined at an angle of 30° to H.P., but which,
with its shortest edges, are to be in vertical planes inclined
to the V.P. at an angle of 30°. Its nearest corner" to be 4"
from v.p.
C to be 4"x4"x8" ; its base to be inclined to the H.P. at an
angle of 45° ; its axis in a vertical plane incHned to V.P.
at an angle of 30° ; and its nearest corner 4" from v. p.
Note. — The axis is a line drawn from the vertex to the centre
of the base ; as ai?, Fig. 250.
D to be 6" X 4" X 4", to have one side inclined at an angle of 30°
with H.P., and its ends in a vertical plane inclined to v.p.
at an angle of 60°. The nearest corner to be 4" from V.P.
Scale ^ full size.
Note. — The heights in these elevations are obtained by first
drawing side views of the objects, as shown in Fig. 249. The
connection is fully shown in Figs. 254 A and B.
SIMPLE SOLIDS.
137
181. To project a quadrilateral prism 6j" x 3^" x 3" resting on one
corner on H.P., and its faces forming equal angles with it, with
its longer edges inclined at an angle of 60° to H.P., and parallel
to v.p. Its nearest edge to be Ij" from v. P. Scale ^ full size.
Fig. 256 A.
Draw XY, and ij" below it draw the line ;;/;;. In any con
venient position draw the line ca perpendicular to ;«;?, and
from c draw c^ 3^" long, at an angle of 45°. From d as centre,
and with radius dc, draw an arc cutting ca in a. Join L?a. Also
draw M perpendicular to ca. This represents onehalf of the
actual shape of the base of the prism.
Fig. 256.
At any point c' on XY draw the line //' 6^" long at an angle
of 60° to XY, and the line /// perpendicular to it. From /,
along e7i', set off the distances e', ^', h\ equal to those of <?, <y, c.
At each of these points draw lines parallel to e'f\ and equal to it.
Join the tops of these lines. This completes the elevation.
Draw lines from b and a parallel to i7in. Every point in the
plan must come on one of these three lines. Drop a line from
/' at right angles to XY till it meets the horizontal line from b \
this gives the point/ Every other point in the plan is found
in the same manner.
f38
GEOMETRICAL DRAWING AND DESIGN.
182. To project the same prism, with its longer edges stUl inclined
at an angle of 60° with H.P., and its faces making equal angles
with it ; taut instead of taeing parallel to v. p., as in Fig. 256 A,
let them be in vertical planes, inclined at an angle of 45° with
v.P. The nearest corner of prism to be Ij" from V.P. Scale J
full size. Fig. 256 B.
Note. — We can always imagine any line to be contained in a
vertical plane, whether the object contains one or not. In Fig.
254 B the line k'o' is contained in the vertical plane J^o'l'g' ; but
in this instance s'w' (Fig. 256 b) is not contained in one, as the
solid does not contain a vertical plane.
At any point o ij" below XY, draw oq at an angle of 45° with
XY. Make oq equal in length to nm (Fig. 256 a). The plan Fig.
2 56 B is precisely similar in shape to the plan Fig. 2 56 A, but turned
to make an angle of 45° with XY ; so if we take the line qo to
represent the line nin^ we can complete the plan from Fig. 256 A.
Every point in the elevation is found in the same way : erect
a perpendicular upon point r till it meets a horizontal hne
drawn from/' (Fig. 256 a) ; this gives the point r', and so on
till the elevation is completed.
183. To project a regular hexagonal prism 10" long, and with sides
3^" wide, standing on its
4r4 taase on H.P., with one
face parallel to v.P. and
2^" from it. Scale jV
full size. Fig. 257 A.
Draw the Hne XY, and
2" below it draw the line
ab. Complete the hexagon.
Above XY draw perpendi
culars 10" long immediately
above the points c^ e^f, d.
Join the tops of these per
pendiculars, which com
pletes the elevation.
184. To project the same prism lying on one face on the H.P.,
with its longer edges parallel to the v.P. ; its nearest edge to
be if" from v.P. Scale ^V full size. Fig. 257 B.
Draw the line /w, 10" long i" below XY and parallel to it.
Draw the lines Is and mp perpendicular to Im. Set off the
SIMPLE SOLIDS.
139
.^ distances w, n^ <?, p on mp equal to the distances c\ /,/', d
I (Fig. 257 a). Draw lines from «, o, and^, parallel to /w, till they
meet the line /s. This completes the plan.
Draw perpendiculars above XY immediately above the points
s, pj and set off the distances u', s', r equal to the distances
K, H, G (Fig. 257 a). Draw lines from the points / and r parallel
to XY till they meet the perpendicular p'd. This completes the
elevation.
185. To project a regular hexagonal prism 7V' long, and with sides
2^" wide, standing on its base on the H.P., with one face in
clined to the v.P.
at an angle of 45". ^' c' ^' g*
Its nearest edge to
be l" from v.P.
Scale \ full size.
Fig. 258 A.
Draw the line XY,
and if" below it mark
the position of point a.
From a draw the
line ab^ i" long, at an
angle of 45° with XY.
Complete the hexagon.
Above XY draw per
pendiculars 7^" long
immediately above the
points b^ r, d^ e, and
join the tops of these Hnes
Fig. 258.
This completes the elevation.
186. To project the same prism, l3ring with one face on H.P. ; its
longer edges to be inclined to the V.P. at an angle of 30° ; its
nearest corner to be l\" from v.P. Scale ^ full size. Fig. 258 B,
Fix the point/ i j" below XY. Draw the line f^, y\" long,
at an angle of 30° with XY ; and from /" and ^ draw lines perpen
dicular to fg. From ^, along the line gk, set off the distances
g, w, /, k, equal to the distances c, i, p, o (Fig. 258 a). From the
points, m, /, and k^ draw lines parallel to fg till they meet the
\\ne.fk. This completes the plan.
140
GEOMETRICAL DRAWING AND DESIGN.
The heights K, H, G, which give the horizontal Hnes in the
elevation, are obtained from the distances 6, z, d in the plan
(Fig. 258 a). Having obtained these heights, draw the lines h'k'
and n'm'. Carry up perpendicular lines from the points in the
plan till they meet these lines, which give the corresponding
points in the elevation. Join them as shown.
187.
To project a regular hexagonal prism 12 V' long, with sides 4"
wide, resting on one of its smaller edges on the h.p. ; its longer
edges to be inclined at an angle of 60° to the H.P., and parallel
to the v.p. Its nearest edge to be 2" from v. p. Scale ^ full
size. Fig. 259 A.
In any convenient position draw the hexagon AEDF with
4" sides, with lines joining the opposite angles as shown. Draw
XY, and at any
point d draw the
line dc\ 12^" long
at an angle of 60
with XY. From a!
and c' draw the
lines ciU and c'd'
perpendicular to
a!c' . Set off the dis
tances a!^ /', b'
along db\ equal to
thedistancesE,B,F
of hexagon. From
these points draw
lines parallel to cic'
till they meet the
line c'd' . This com
pletes the elevation.
Draw a line 2" below XY, and parallel to it. From L on
this line let fall a perpendicular, and on it set off the distances
L, K, H, G equal to the distances A, B, C, D of hexagon. From
each of these points draw lines parallel to XY, and let fall
lines from the various points in the elevation till they meet
these lines. This gives the corresponding points in the plan.
Join them as shown.
259
SIMPLE SOLIDS.
141
188. To project the same prism standing on one of its shorter
edges on the H.P., at an angle of 30^ with v. p., with its longer
edges inclined at an angle of 60° with the H.P., and in vertical
planes inclined at an angle of 60° to v.p. ; its nearest comer to
be 3" from v.p. Scale ^V full size. Fig. 259 B.
Fix the position of point vi 3" below XY, and draw the line
vip at an angle of 30° with XY. This represents the line KH in
Fig. 259 A. The plan in Fig. 259 B is precisely the same as that
shown in Fig. 259 A, turned to a different angle. Complete the
plan against the line ;;//, from Fig. 259 A. As every point in the
elevation is found in the same way, it is only necessary to
describe one point. Erect a line on point r at right angles
to XY till it meets a horizontal line drawn from point e' . This
gives the point r. Find the other points in the same way, and
join them, as shown.
189. To project a regular hexagonal prism, Tj" long, with faces 2V
wide, resting on one comer on the H.P. ; its longer edges to be
inclined at an angle of 45° with the H.P.; one face to be parallel
to the v.p. and 2V' from it. Scale \ full size. Fig. 260 .A..
Construct a hexagon with 2o" sides, and join the opposite
J?
, d.'
k
angles by lines at right angles to each other, as shown. Draw
the line XY, and at any point (^ draw the line a!h\ 7\" long, at
an angle of 45° with XY. From the points a' and h' draw the
142 GEOMETRICAL DRAWING AND DESIGN.
lines rtV and Hd' perpendicular to cih' . Set off the distances
6', e\ f\ d on h'd' equal to A, B, C, D of hexagon, and draw lines
from these points parallel to h'a' till they meet the line c^c.
This completes the elevation.
Draw the line mk 2V below XY. From H on mk produced
draw the line HL perpendicular to mk, and set off the distances
H, D, L equal to the distances E, B, F of hexagon. Draw Hnes
from these points parallel to XY. All the points of the plan
must come on these three lines, and are determined by dropping
perpendiculars from the corresponding points in the elevation.
190. To project the same solid, with its longer edges still inclined at
an angle of 45° to the H.P. ; but instead of being parallel to the
V.P., as in the last problem, let them be in vertical planes,
inclined at an angle of 30° with the v. P. ; its nearest corner to
the v.p. to be Ij" from it. Scale ^ full size. Fig. 260 B,
Fix the position of point n ij" below XY, and draw the line
no at an angle of 30° with it. The plan in this problem is
precisely similar to the plan in the last problem, but turned round
at an angle of 30° with XY, and the line no corresponds with
the line km (Fig. 260 a). Complete the plan as shown.
Every point in the elevation is found in the same manner.
For example, erect a perpendicular on point o till it meets a
horizontal line drawn from point e'; this will give the point o'.
Proceed in the same way with all the other points, and join them.
The Regrular Solids.
There are five regular solids, and they are named after the
number of faces they each possess ; viz. the Tetrahedron has four
faces, the Hexahedron six faces, the Octahedron eight faces, the
Dodecahedron twelve faces, and the Icosahedron twenty faces.
They possess the following properties, viz. :
(i) The faces of each solid are equal, and similar in shape,
and its edges are of equal length. (2) All their faces are regular
polygons. (3) All the angles " formed by the contiguous faces
of each solid are equal. (4) Each can be inscribed in a sphere,
so that all its angular points lie on the surface of the sphere.
The student is advised to make these drawings to a scale
three times that of the diagrams.
SIMPLE SOLIDS.
H3
191. To project a tetrahedron with edges 9^' long, with one face
resting on the H.P. ; one of its edges to be at an angle of 16°
with V.P., and its nearest angular point 3" from v.P. Scale
jV full size. Fig. 261 A.
All the faces of this solid are equilateral triangles.
Draw XY, and fix the position of point a 3I" below it. From
a draw the line ah
9" long, at an angle
of 16° with XY. On
the line ab con
struct an equilateral
triangle ahc. Bisect
each of the angles
at ^, 6, and c by
lines meeting at d.
This completes the
plan.
To find the alti
tude of the eleva
tion, produce the
line bd to e, and at
d draw the line df
perpendicular to eh.
With b as centre, and radius be, draw an arc cutting d/mf.
To draw the elevation, erect a perpendicular d'g' directly
above d, and make d'g' equal to df. Carry up the points a, c, b till
they meet XY in a', c', and b'. Join the point d' to a'^ c , and b'.
Fig. 261.
192. To project the same solid tilted on to one of its angular points
with its base inclined at an angle of 20^ with H.P. Scale ^V full
size. Fig. 261 B.
Fix the point k' on XY, and draw the line k'k' at an angle
of 20° with it, and equal in length to the line b'a' (Fig. 261 a). As
this elevation is precisely similar to the last, complete it from
that figure on the line k'k'.
Draw lines from the points a, b, d, and c (Fig. 261 a) parallel
to XY. All the points of the plan must come on these four
lines, and are found by dropping lines at right angles to XY
from the corresponding points in the elevation.
144
■GEOMETRICAL DRAWING AND DESIGN.
193. To project a hexahedron or cube with edges 3" long, resting
on one edge on the H.P., and its base making an angle of 22°
with it; the side nearest the V.P. to be parallel to, and l"
from it. Scale ^ full size. Fig. 262 A.
All the faces of this solid are squares.
Draw the line XY, and at any point d draw the line cih' at an
angle of 22°. Complete the square dh'c'd' .
Fig. 26
Below XY draw the line ef i" from it. Draw the lines a'a^ b'b,
c'c and d'd at right angles to XY. Make eb and fd each equal
to dh\ and join hd \ cc' cuts efm.g.
194. To project the same solid with its edge stiU resting on the H.P.
but inclined to the v. P. at an angle of 60°, the base still forming
the same angle with the H.P., Tdz. 2°; its nearest corner to be
1' from V.P. Scale \ full size. Fig. 262 B.
Fix the position of point h \" below XY, and draw the line hk
equal X.o fe (Fig. 262 a) at an angle of 30° with XY. Complete
this plan from Fig. 262 A, to which it is precisely similar in shape.
Draw lines through the points 5', </, and d (Fig. 262 a) parallel
to XY. Draw perpendiculars from the points in the plan till
they meet these lines, the intersections give the corresponding
points in the elevation. Join these points as shown.
SIMPLE SOLIDS.
HS
195. To project an octahedron, with edges 4" long, poised on one of
its angiUar points on the H.P.; its axis to be perpendicular to
H.P., and its edge nearest to the v.P. to be parallel to, and l"
from it. Scale ^ full size. Fig. 263 A.
All the faces of this solid are equilateral triangles.
Draw the line XY, and if" below it draw the line ab 4" long.
Complete the square ^
ab(/c, and draw its di
agonals. This will be
the plan of the solid.
As every angular
point is equidistant
from the centre of a
sphere circumscribing
the solid, the point e
must be the centre of
the plan of the sphere
and (u/ its diameter.
Draw the projector
ee' from the point t\
Make e'f equal in
length to ac/. This will
be the axis of the solid.
Bisect e'/' by the line cUf parallel to XY, and erect perpendi
culars from the points c and ^/till they meet this line in the points
c' and d. Join c'e' and c'f, d'e and d'f\ to complete the elevation.
196. To project the same solid with one face resting on the H.P. ;
the edge nearest the v.P. to be parallel to, and l" from it.
Scale \ full size. Fig. 263 B.
Fix the positions of the points g' and // on XY the same
distance apart as the points c' and/' in the preceding problem.
On this line complete the elevation from Fig. 263 A.
Draw lines from the points 6, e, and d parallel to XY.
All the points of the plan must come on these three lines,
and are found by dropping perpendiculars from their correspond
ing points in the elevation.
The Dodecahedron and Icosahedron can hardly be described
as ' simple solids.' Their projection will be found in Spanton's
Complete Geometrical Course {Macinilla?i\ pp. 229 sqq.
K
146
GEOMETRICAL DRAWING AND DESIGN.
Octagonal Pyramids.
197. To project a re^lar octagonal pyramid, 8" high, with each
side of its base 2h" wide, standing on its base on the H.P., with an
edge of its base parallel to the V.P. and 2^" from it. Scale jV full
size. Fig. 264 A.
Draw XY, and 2h" below it draw ad 2\" long. On ab
construct a regular octagon, and join the opposite angles
Carry up projectors from the points d, e^f^g perpendicular to
XY. Fix the point c' immediately above r, and 8" above
XY. Join the point c' to d'^ e\f\ and g\ which completes the
pyramid.
198. To project the same solid lying with one face on the H.P., but
with its axis in a plane parallel to the v. P. Scale ^V full size.
Fig. 264 B.
On XY mark off the distance J^k' equal to g'c' (Fig. 264 a).
Complete the construction of elevation from Fig. 264 A.
Let fall lines from the various points of the elevation, at right
angles to XY, till they meet lines drawn from the corresponding
points in the plan (Fig. 264 a), parallel to XY. The intersection of
these lines give the required points, by joining which we obtain
the plan.
SIMPLE SOLIDS.
147
199. To project the same solid resting on one of its shorter edges,
■with its base inclined at an angle of 30^ with H.P. ; its axis to be
in a plane parallel to the v. P. Scale ^ full size. Fig. 264 c.
Draw XY, and fix the position of point o'. Draw o'p' at an
angle of 30° with XY, and equal in length to /I'n' (Fig. 264 b).
Complete the elevation from Fig. 264 b.
In any convenient position draw the line DG perpendicular to
XY, and set off upon it the distances D, E, F, G equal to /i\ /',
m', n' (Fig. 264 b). Draw lines from these points parallel to XY.
From the various points in the elevation let fall lines at right
angles to XY till they meet these'lines, which give the correspond
ing points in the plan.
Cones.
Cones are projected in precisely the same way as polygonal
pyramids. After finding the points in the base, instead of
joining them by lines, as in the case of pyramids, a fair curve is
drawn through them. Eight points of the base are found in the
examples here given. Should more points be required, it is
only necessary to select a pyramid having more sides than
eight to construct the cone upon,
200. To project a cone 8" high, with base 6h' in diameter, resting
on the H.P. Scale yV full size. Fig. 265 A.
Draw XY, and in any convenient position below it draw a
circle 6V' in diameter, and a diameter ^/i parallel to XY.
1' .V'
Carry up projectors from g and /i till they meet XY ; also
from k, and produce ke 8" above XY. Join k'g' and ^7i\
148 GEOMETRICAL DRAWING AND DESIGN.
201. To project the same solid resting on its edge, with its base
inclined at an angle of 30^ with H.P. ; its axis to be in a vertical
plane parallel to the V.P. Scale ^V fuU size. Fig. 265 B.
Inscribe the circle (Fig. 265 a) in a square abcd^ the side
nearest XY to be parallel to it. Draw diagonals, and through
the centre draw diameters parallel to the sides of the square.
Through the four points where the diagonals cut the circle
draw lines parallel to the sides of the square.
Fix the point m' on XY, and draw the line m'r' at an angle
of 30° with it. Set off the distances m'o'p'q'r' equal to the dis
tances /?', ;/, e\ l\g' (Fig. 265 a). Complete the elevation from
Fig. 265 A.
Let fall lines from the various points of the elevation at
right angles to XY till they meet lines drawn from the corre
sponding points of the plan (Fig. 265 a). Draw a fair curve
through the points forming the base, and lines from the vertex s
tangential to the base. This completes the plan.
202. To project the same solid, resting on its edge, with its base
still inclined at an angle of 30° with H.P., but with its axis in a
vertical plane, inclined at an angle of 60° with V.P, Scale j^Tr full
size. Fig. 265 C.
Draw the lines enclosing the base with the parallel lines
intersecting each other where the diagonals cut the circle from
the plan (Fig. 265 b). The line xy that passes through the axis
to be inclined at an angle of 60° with XY. Complete the plan
from Fig. 265 B.
Draw lines from the various points of the plan at right
angles to XY till they meet lines drawn from the corresponding
points of the elevation (Fig. 265 b). Complete the elevation as
shown.
Cylinders.
In the examples here given, only eight points of the circular
base are projected, to save confusion of lines ; but any number
of points can be found in the same manner.
SIMPLE SOLIDS.
149
203. To project a cylinder, 5^" in diameter and 8j" high, standing
on its base on the h.p. Scale ^ full size. Fig. 266 A.
Draw XY, and in any convenient position below it draw a
circle 5V' in diameter. Draw tangents to it at the points aandd
perpendicular to XY,
and produce them 8j"
above XY. Join the
tops of these lines to
complete the cylinder.
204. To project the
same cylinder, ly
ing on its side on
the H.P., with its
axis inclined at an
angle of 45° with
the V.P. Scale ^
full size. Fig. 266B.
Draw four diameters
tothe plan (Fig. 266 a),
by first drawing the
line ad parallel to XY,
and then the other three diameters equidistant from it. This
gives eight points in the circumference. Draw lines from these
points parallel to XY, till they meet the line AE in the points
A, B, C, D, E.
At any convenient point c below XY (Fig. 266 b), draw the
line cf at an angle of 45° with it. Draw the line cm, 8" long,
perpendicular to c/, and from m draw m£ parallel to cf. From
m, along 7n^, set off the distances w, /, >^, //,^ equal to the distances
A, B, C, D, E (Fig. 266 a), and from these points draw hues parallel
to c?n. This completes the plan.
Draw a line from / perpendicular to XY, and produce it
above the ground line. Set off the distances E', D', C, B', A' on
this line, equal to the distances E, D, C, B, A (Fig. 266 a), and
draw lines from these points parallel to XY. Draw the pro
jectors from the various points in the plan till they meet these
lines, which give projections of the points required to complete
the elevation.
Fig. 266.
ISO
GEOMETRICAL DRAWING AND DESIGN.
205. To project the same solid, resting on its edge, with its base
inclined at an angle of 30° with the H.P., its axis being parallel
to the v.p. Scale ^ full size. Fig. 267 A.
Draw XY, and at any point n' upon it draw the line n'r'
at an angle of 30° with the ground line. On the line n'r' set
off the distances n', o', p\ q\ r' equal to the distances g^ h, k. /, m
(Fig. 266 b), and from each of these points draw perpendiculars
to n'r\ 8j" long. Join s'^u'. This completes the elevation.
From y let fall a line at right angles to XY, and set off upon
it from any convenient point A the points B,C,D, E equal to the
distances n\ o\ p\ q\ r'. From each of these points draw lines
parallel to XY. From the various points in the elevation drop
projectors till they meet these Hues in the corresponding points,
by connecting which we get the plan.
206. To project the same solid resting on its edge, with its base still
inclined at an angle of 30° with the H.P., but with its axis inclined
at an angle of 60° with the v.p. Scale ^ full size. Fig. 267 B.
Draw the line FG equal to AE, inclined at an angle of 30°
with XY. Complete the plan from Fig. 267 A. Draw projectors
from the plan till they meet lines drawn parallel to XY from
the corresponding points in the elevation (Fig. 267 a). These
give the necessary projections for completing the elevation.
SIMPLE SOLIDS.
151
Spheres.
The plan and elevation of a sphere are circles ; but if we
divide the sphere into divisions by lines upon its surface, such
as meridians of longitude and parallels of latitude, we shall be
enabled to fix its position and inclination to the coordinate
planes, and project it accordingly.
So as not to confuse the figure too much, we will restrict
ourselves to eight meridians, with the equator, and two parallels
of latitude. The junction of the meridians will of course give
us the position of the poles, which will determine the axis.
207. To project a sphere 5^" in diameter, with meridians and
parallels ; its axis to be perpendicular to the H.P.
size. Fig. 268 A.
Draw XY, and
in any conven
ient position be
low it draw a
circle 5^" in dia
meter. Draw the
diameters ab
parallel to the
ground line, and
rtf/at right angles
to it, and two
other diameters
equidistant from
them.
Produce the
line dt above
XY, and make
c'p' equal in
length to the diameter ab. Bisect c'p' in d' . With d as centre,
and radius equal X.opa, draw a circle. Draw db' through d till it
meets the circle in d and b' . Through d' draw the line e'f at
an angle of 45° with db'., till it meets the circle in e and /'.
From e' and /' draw lines parallel to db' till they meet the
circle in g' and //. These lines represent parallels of latitude.
Drop a perpendicular from g' till it meets ab in g. With p as
Fig. 268.
152
GEOMETRICAL DRAWING AND DESIGN.
centre, and radius pg^ draw a circle,
parallel c'g' .
v'
This is the plan of the
To avoid con
fusion, the pro
jectors for half
of one meridian
only are shown ;
but they are all
found in the
same manner.
Erect a per
pendicular on
point k till it
meets the equa
tor in point k' ;
also from point
/till it meets the
parallels in
points /" and /'.
Draw a curve
which gives the projection of
through the points c\ /", k\ /', p\
the meridian.
208. To project the same sphere, with its axis inclined to the H.P.
at an angle of 60°, but parallel to the V.P. Scale \ full size.
Fig. 268 B.
Note. — The same letters are taken throughout these spherical
problems to facilitate reference.
Draw a circle 5^" in diameter, touching XY, and draw the
line c'p' at an angle of 60" with it. Draw the line db' at right
angles to c'p\ and set off the distances of the parallels above and
below a!b' equal to their distances in the elevation (Fig. 268 A).
Draw the lines e'g and hf parallel to alb'. Complete the
elevation from Fig. 268 A. Draw lines from all the points of
intersection between the meridians and parallels of the plan
(Fig. 268 a) parallel to XY, and let fall perpendiculars from the
corresponding points in the elevation till they meet these lines,
which give the projections of the points of intersection. Draw
the curves.
SIMPLE SOLIDS.
153
209. To project the same sphere, with its axis still inclined to the
H.P., at an angle of 60°, hut in a
vertical plane inclined at an angle of
60° with the V.P. Scale ^ full size.
Fig. 269.
Draw the line cp at an angle of 60°
with XY, for the plan of the axis, and on
this line complete the plan from Fig. 268 B.
Perpendicular to XY draw the line XL,
and set off the distances X, C, L, K, C, L
equal to the distances Y, C, L, K, C, L
(Fig. 268 b). From each of these points
draw lines parallel to XY till they meet
projectors drawn from the corresponding
points in the plan, which give the projec
tions required.
Fig. 269.
EXERCISES.
1. The plan is shown of three bricks (Fig. 270), each 9" x 4^" x 3",
one resting upon the other two. Draw an elevation upon the given
line xy.
Scale (which need not be drawn) 2" to i', or \ of full size. (April, '98.)
Fig. 270.
Fig. 271
2. The plan is given (Fig. 271) of a flight of three steps each "
high, of which 3 is uppermost. Draw an elevation on the given xy.
(June, '00.)
J,54
GEOMETRICAL DRAWING AND DESIGN.
3. Plan and elevation are given of a solid letter H (Fig. 272). Draw
an elevation when the horizontal edge ab makes an angle of 60 with the
vertical plane of projection. (June, '97.)
G, b
A
Fig. 272. Fig. 273.
4. The diagram ( Fig. 273) shows the plan of two square prisms, one
resting upon the other. Draw their elevation upon the given xy. The
lines AB and CD are plans of square surfaces. (June, '99.)
5. The plan is given of a cube (Fig, 274), having a cylindrical hole
pierced through its centre. A vertical plane, represented by the line Im,
+
IlevcUion
Fig. 274. Fig. 275.
cuts off a portion of the solid. Draw an elevation on the line xy, supposing
the part of the solid in front of Im to be re
moved. The part in section should be clearly
indicated by lightly shading it. (April, '96.)
6. The plan is given (Fig. 275) of a right
prism having equilateral triangles for its bases.
These bases are vertical. Draw an elevation
of the prism on the line xy. Show the form
of the section of the prism made by the vertical
plane Im. The part in section should be indi
cated by lightly shading it. (June, '98.)
7. Plan and elevation are given (Fig. 276)
of a solid composed of a halfcylinder placed
f'ig 276. upon a prism. Draw a new elevation, when
the horizontal edges of the prism make angles of 45° with the vertical
plane of projection. (April, '96. )
SIMPLE SOLIDS.
^55
8. An elevation is given (Fig. 277) of an archway with semicircular
head, in a wall i' 6" thick. Draw a second elevation upon a vertical
plane which makes an angle of 45° with the face of the wall. (Scale,
which need not be drawn, ^' to i'.) (June, '00.)
Fig. 277.
Fig. 278.
Fig. 279
9. Plan and elevation are given (Fig. 278) of a rectangular block with a
semicylindrical hollow in it. Draw a new elevation upon a vertical plane
which makes an angle of 30° with the horizontal edge ab. (April, '98.)
10. The diagram (Fig. 279) shows a side elevation of a square prism,
pierced through its centre by a cylinder. Draw a front elevation of
the solids. (June, 'cxd. )
Elevation
PlOJl
Fig. 280.
Fig. 281.
11. Plan and elevation are given (Fig. 280) of a cylinder through
which a square opening has been cut. Draw a fresh plan and elevation
of the solid, the plane of the circular base FG being inclined at 45° to
the vertical plane of projection. (June, '99.)
12. The "block" plan and end elevation are given (Fig. 281) of a
building having a square tower with pyramidal roof. A side elevation
of the building is required. (June, '98.)
156
GEOMETRICAL DRAWING AND DESIGN.
13. The plan is given of a piece of cylindrical rod (Fig. 282) cut by a
vertical plane shown at Im. Draw an elevation of the solid upon an xy
parallel to Im. (April, '99.)
Fig. 282.
Fig. 283,
14. Plan and elevation are given of a sloping desk (Fig. 283). Draw
an elevation upon a vertical plane parallel to the Xvtxo. pq. Show upon
this elevation the outline of the section made by the vertical plane
represented by /</. (April, '99.)
15. The diagram (Fig. 284) shows the plan of two cubes, one resting
upon the two others, with a sphere resting on the upper cube. Draw
an elevation on the given xy. (April, '00.)
Fig. 284.
Fig. 285.
16. Show in plan and elevation a shallow circular metal bath.
Diameter at top 2' 6", at bottom 2', height 6". The thickness of the
metal may be neglected. Scale (which need not be drawn) i' to \".
(April, '99.)
17. The diagram (Fig. 285) shows the elevation of a right cone having
its vertex at V. Draw the plan. (April, '00.)
CHAPTER XIV.
ORTHOGRAPHIC PROJECTION.
Preparatory to the study of sections of solids it is desirable
to have a more thorough insight into the principles of Ortho
graphic Projection, though its simpler applications need only be
considered.
In Solid Geometry objects are projected by means of parallel
projectors perpendicular to
two coordinate planes. These
planes may be considered as in
definite in extent. For instance,
the H.P. might be extended
beyond the V.P., and the v. p.
below the H.P.
To understand this fully, let
us take two pieces of cardboard
about 12" square, and halfway
across the middle of each cut a
groove, as shown in Fig. 286.
By fitting these two pieces
together we obtain two planes intersecting each other at right
angles, as shown in Fig. 287.
We have now four sets of coordinate planes, forming four
" dihedral angles," identified by the letters A, B, C, D.
The angle formed by the upper surface of the H.P. with the
front of the V.P. is called the " first dihedral angle," viz. A, fig. 287.
The angle formed by the upper surface of the H.P. with the
back of the V.P. is called the " second dihedral angle," viz. B, fig. 287.
The angle formed by the under surface of the H.P. with the
back of the v.p. is called the "third dihedral angle," viz. C, fig. 287.
The angle formed by the under surface of the H.P. with the
front of the V.P. is called the " fourth dihedral angle," viz. D, fig. 287.
158
GEOMETRICAL DRAWING AND DESIGN.
We will now take a piece of cardboard 4" x 3", and place one
of its shorter edges against the H.P. and a longer edge against
the v.p. in the first
dihedral angle, with
its surface perpen
dicular to each
plane (Fig. 287).
Let the corner A
represent a point
we wish to project
on to each plane :
the top edge Aa'
represents its verti
cal projector, and
the point a! its ver
tical projection; the
edge Ka represents
its horizontal projector, and the point a its horizontal projection,
and so on, placing the cardboard in each of the dihedral angles.
Lines.
To illustrate the projection of lines, we will restrict ourselves to
the two coordinate planes
of the first dihedral angle
(Fig. 288).
Take the piece of card
board and place it with
one of its shorter edges
on the H.P., with its sur
face parallel to the v.p.
Let the top edge AB
represent a line we wish
to project. The edges ha
and B6 will then represent the horizontal projectors, and the line
ah its horizontal projection. If we draw lines A«' and B6'
perpendicular to the V.P. from the points A and B, they represent
the vertical projectors, and the line clU its vertical projection.
We will now place the piece of cardboard touching both
planes, with one of its shorter edges on the H.P., and its surface
ORTHOGRAPHIC PROJECTION,
r?9
perpendicular to both planes. Let the edge Cc represent the
line to be projected, Cc" and cc' represent the vertical projectors,
and the line c'c" its vertical projection. The point c on the H.P.
is called the " horizontal trace " of the line.
Note. — The point where a line, or a line produced, would
meet either plane is called the " trace " of that line : if this point
is on the H.P., it is called the "horizontal trace" (h.t.); and if it
is on the v.p., the "vertical trace" (v.T.). The same thing
applies to the projection of planes.
Now place the piece of cardboard with one of its longer edges
on the H.P., and its surface per
pendicular to both planes. Let
the top edge Dd" represent the
line to be projected. The edges
D^and ^W represent the hori
zontal projectors, and the line X
dd' its horizontal projection.
The point d" is its vertical trace.
Fig. 289 represents the co
ordinate planes opened out
into one flat surface. The pro
jections below XY represent Fig. 289.
the plans of the lines, and those above XY the elevations.
We will now use the same piece of cardboard to illustrate the
projections of lines inclined to one or both coordinate planes
a
b
,c '
i /
■
\c'
d'
b
i
a
^0
d
Fig. 290.
(Fig. 290). In the first case we will incline it to both planes, with
one of its shorter edges resting on the H.P. and parallel to the v,P
i6o
GEOMETRICAL DRAWING AND DESIGN.
Let the edge AB represent the Hne to be projected. aB is its
horizontal, and a'd' its vertical projections.
Then incline it to the H.P., with one of its shorter edges still
on the H.P., but perpendicular to the v.P.
Let CD represent the line to be projected. The line D^: is
the horizontal, and c'ci' its vertical projections.
Now incline it to the V.P., with its lower longer edge parallel
to, but raised a little above the H.P,
Let EF represent the line to be projected. The line e/ is its
horizontal, and/V its vertical projections.
Let us now draw a diagonal GF across the piece of cardboard
and again hold it in the same position ; and let GF represent
the line to be projected.
The line e/ still represents its horizontal projection, but the
line/'^' is its projection on the vertical plane.
Fig. 291 shows the plans and elevations of these Hnes, with the
coordinate planes opened out flat.
We have now projected a line in seven distinct positions,
viz. :
Fig. 288. — AB parallel to H.P. and parallel to
C perpendicular to „ „
D parallel to ,
Fig. 290.
AB inclined to
CD
EF parallel to
GF inchned to
a'
/
c '
/
/ 1 '
b'
V
a
/
R
i
/
Pi
' f
/
Fig. 291.
perpendicular to
inclined to
parallel to
inclined to
The student should parti
cularly notice the difference
between AB and FG in
Fig. 290. Although they
are both inclined to both
planes, AB is in a vertical
plane perpendicular to the
V.P., while FG is in one
inclined to the v.p.
We will now project these
lines in the various positions
to scale.
ORTHOGRAPHIC PROTECTION.
i6i
210. To project a line AB 2^" long, parallel to botli the H.P. and
V.P., its distances to be 3" from the H.P. and ih" from, the V.P.
Scale J full size. Fig. 292.
Draw XY, and iV' below it draw the line a^ 2^' long. Draw
the projectors aa' and bb' at right angles to XY, and 3" above it.
Join a'b'.
211. To project a line CD 3" long to the same Scale, parallel to
the V.P. and 2^" from it, but perpendicular to the H.P. Fig. 292.
Fix the position of point c 2^" below XY. Draw a line per
pendicular to XY, and produce the same 3I" above it. c is the
plan or H. trace, and c'd' the elevation required.
212. To project a line EF 3" long to the same scale, parallel to the
H.P. and 2j" above it, but perpendicular to the v. p. Fig. 292.
Below XY, and perpendicular to it, draw the line ef 3" long.
Draw the projector /V 2^" long, ef is the plan, and e' the
elevation or V. trace.
213. To project a line GH 3" long to the same scale, parallel to the
V.P. and 1^" from it, but inclined to the H.P. at an angle of 60°o
Fig. 292. "
At any point £' on XY draw the line g^/t' 3" long, and inclined
to the H.P. at an angle of 60°. Let fall the projectors o^'o and
A'A at right angles to XY. Set off the points ^ and /i i^" below
XY, and join them.
214. To project a line KL 3 " long to the same scale, inclined to the
H.P. at an angle of 60°, but in a vertical plane perpendicular to
the V.P. Fig. 292.
Draw the line k7 at right angles to XY till it meets horizontal
Hnes drawn from // and //. k'l' is the elevation, and /'/the plan.
L
J62
GEOMETRICAL DRAWING AND DESIGN.
215. To project a line MN 3" long to the same scale, parallel to the
H.P. and l" atoove it, but inclined to tlie v. p. at an angle of
45°. The end of the line nearer the V.P. to be " from it.
Fig. 292.
Fix the point n " below XY, and draw
5" long at an
angle of 45° with it. Carry up the projectors perpendicular to
XY, and produce them i^" above it in the points in' and ;/.
Join in'7i' .
216. To project a line OP 3" long to the same scale, inclined to the
H.P. at an angle of 30°, but in a vertical plane inclined to the
V.P. at an angle of 60 ; one end of the line to be on XY.
Fig. 292.
From point o' on XY draw a line o'K 3" long, and inclined
to XY at an angle of 60°. From the same point o' draw the
line ^'B at an angle of 30° with XY. With o' as centre, and
radius <?'A, draw an arc till it meets ^'B in B. Draw the Hne
BC perpendicular to XY. With 0' as centre, and radius o'C^
draw an arc till it meets o' k. mp. Draw the projector pp' till
it meets a horizontal line drawn from B in p' . Join p'o'. op
is the plan, and op' the elevation of the line required.
Planes.
The lines in which planes intersect the coordinate planes
are called traces : if on the H.P., the horizontal trace (h.t.) :
Fig. 293.
and on the v.p., the vertical trace (v.t.). The inclination of
planes is determined by means of these traces.
ORTHOGRAPHIC PROJECTION.
163
We will take the same piece of cardboard that we have used
for our previous illustrations and place it on the H.P. and
parallel to the V.P., as A (Fig. 293), The line ab^ where it
intersects the H.P., will be its H.T.
If we place it parallel to the H.P. and perpendicular to the v. p.,
as B, the line </</, where
it intersects the v. p.,
will be its v.t.
By placing it perpen
dicular to each plane, as
C, ^will be its H.T. and X
eg its V.T.
On opening these co
ordinate planes out flat
these traces will appear
as shown in Fig. 294.
We will now place the
piece of cardboard perpendicular to the H.P. and inclined to
the V.P., as D (Fig. 295) : hk will then be the H.T., and ///'
the V.T.
Fig. 295.
Now incline it to the H.P. and make it perpendicular to the
V.P., as E (Fig. 295) : mn will be its H.T. and mo' its V.T.
By inclining it to both planes with its shorter edges parallel
to XY, as F (Fig. 295), pq will be the H.T. and r's' the V.T.
164
GEOMETRICAL DRAWING AND DESIGN.
For our next illustration we will take a 60° set square G, as
a right angle will not fit closely to the two planes in this posi
tion ; tu will be the H.T. and tw' the V.T.
If we now open the planes as before, these traces will be
shown as in Fig. 296.
,
0'
r' s'
/o^
w'
/
t^
/
/
aW
m
t^
^
/
5
1
■) q
\
Fig. 296.
From these illustrations we can 'deduce the following facts —
A plane can have no trace on the plane it is parallel to (see
A and B, Fig. 293.
If traces are not parallel to XY, they must intersect each
other on that line (see D, E, and G, Fig. 295).
If the traces of a plane are in one straight line when the H.P.
and v.P. are opened out so as to form one continuous surface,
the angles the plane forms with each coordinate plane must
be equal.
When a plane is perpendicular to either coordinate plane,
its inclined trace will always give the amount of its inclination
to the other coordinate plane ; e.g. hk (Fig. 296) forms with XY
the angle ^ or the inclination of D (Fig. 295) to the v.P., while
mo' forms with XY the angle d or the inclination of E to the H.P.
When a plane is inclined to both planes, but has its traces
parallel to XY, the sum of its inclinations, i.e. d+cfi = go°; as
F (Fig. 295).
The traces of planes inclined to one or both planes are not
supposed to finish at XY ; they are indefinite, and are generally
produced a little beyond XY.
ORTHOGRAPHIC PROJECTION.
165
217. To find the traces of the foUowing planes. Scale ^ full size.
Fig. 297.
A,3" X 2^" perpendicular to the H.P. and inclined to the v.P. at 60°.
B,3rxif „ „ v.P. „ „ H.P. „ 45°
C,4^"X3" inclined „ H.P. at 60° with shorter edges
parallel to each plane.
Draw XY ; and at any convenient point a draw a6 3" long,
and at an angle of 60° with XY. From a draw ac' 2^" long.
Then ad is the H.T., and ac' the v.T. of A.
Fig. 297.
From any convenient point </ draw ^/' 3I" long, and at an
angle of 45° with XY. From d draw de if" long perpendicular
to xy. Then de is the H.T., and t/f the V.T. of B.
Take any point ^ on XY, and draw o/i' 4I" long at an angle
of 60° with it. From // draw /I'k perpendicular to XY ; and
from o as centre, with radius ^k, draw an arc till it meets a
perpendicular from o in /. From g draw a perpendicular till it
meets a horizontal line from // in 7/1'. Draw ;;/'// and /o, each
3" long, parallel to XY. Then 7/1' 72' will be the v.T., and 0/ the
H.T. of C.
We will now proceed with planes that are inclined to both
planes of projection : they are called oblique planes. Let us take
1 66
GEOMETRICAL DRAWING AND DESIGN.
a 60° set square and place it so as to fit closely against both
planes, as shown at A (Fig. 298). ca will be the H.T. and cU
the v.T.
The inclination of a plane to the coordinate plane containing
its trace is the angle between two lines perpendicular to the
trace, one in the coordinate plane and one in the plane itself.
Fig. 298.
The line ca (Fig. 298) is the H.T. of the plane A, and ab' is
a line in the plane A, and af a hne in the H.P., both perpen
dicular to the H.T.; therefore b'af is the angle A forms with
the H.P.
218. To determine the traces of a plane inclined at an angle of 45°
to the H.P., and at an angle of 35° to the v.P. Fig. 299.
Note. — In the Definitions (page 10) a cone is described as
being generated by the revolution of a rightangled triangle
about one of its sides as an axis. The hypotenuse of this
triangle is called a generatrix.
The problem is generally solved in the following manner :
The generatrices of two cones forming the necessary angles to
the two planes of projection are determined with their axes
meeting at the same point on XY. The sides of these two
cones should be tangential to a sphere, the centre of which is the
ORTHOGRAPHIC PROJECTION. 167
point on XY in which their axes meet. The plane required is
tangential to the bases of these two cones.
Draw XY. Select any point c for the point in which the
axes of the cones meet, and draw a line through it at right
angles to XY, At any point d on XY draw a line at an angle
of 60° with it till it meets the perpendicular on c in b' . With c
as centre, and radius cd^ draw the semicircle def. Join fb'.
Then def\% the plan, and /^W the elevation of a semicone.
From c draw the line eg perpendicular to dU . With c as
centre, and radius cg'^ draw a circle. This will represent the
plan and elevation of a quarter of the enveloped sphere.
Draw the line ah^ at an angle of 45°, tangential to the
plan of the sphere, cutting b'c produced in a. With c as
centre, and radius ch^ draw the semicircle hk'l. Join al.
Then hal will be the plan, and Ik'h the elevation of another
semicone.
From a draw the line m7i tangential to the semicircle y^^;
i.e. the base of the horizontal semicone.
From b' draw the line b'ln tangential to the semicircle Ik'h \
i.e. the base of the vertical semicone.
Then a7n is the H.T. and b'm the V.T. required.
i68 GEOMETRICAL DRAWING AND DESIGN.
Fig. 300 is a perspective view showing this construction. The
horizontal semicone is dotted in each instance.
There is another method of finding the traces for an oblique
plane, viz. by first finding the projections of a line perpendicular
Fig. 300.
to the plane required, and then drawing the traces at right
angles to these projections. This will be more easily understood
by referring to the setsquare B (Fig. 298).
Let Op represent a line at right angles to on\ and perpen
dicular to the plane B. Then op is the horizontal projection,
and o'p the vertical projection of this line {Op) ; and the H.T. ;;?<?,
and the V.T. inti ^ are at right angles to these two projections.
{For Exercises see p. 226,)
CHAPTER XV.
SECTIONS OF SOLIDS, CONSTRUCTION OF SECTIONAL
AREAS.
A section is defined as the intersection of a solid by a plane.
This plane is called the cutting plane, and in the following
problems it is given inclined at different angles to both the
coordinate planes. The surface of the solids cut through are
projected, and the true shapes of the sectional area are
" constructed."
219. To project a cube of V edge, standing on the H.P., and inclined
at an angle of 30 to the v. P., intersected by a cutting plane
inclined to the H. P. at an
angle of 45", and perpendi
cular to the V.P. ; the plane
to intersect both the hori
zontal faces of the cube.
Fig. 301.
Draw the plan adcd of the
cube, and carry up projectors
from the points, " above XY,
and join them for the elevation.
Find the traces of the cutting
plane (Prob. 218).
Where the v.T. cuts the ele
vation in the points e' and /',
drop projectors which will in
tersect the plan in the lines e/i '
and /£. afgche is the plan of
the cut surface of the cube, and e'a'c'f the elevation.
I70
GEOMETRICAL DRAWING AND DESIGN.
The sectional surface can be " constructed ■' by rotating thf
projecting surface of the section on either the h.t. or V.T.
To rotate the plan of the section on the H.T. Draw lines at
right angles to H.T. from the points of the plan, and make the
lengths of these lines from the H.T. equal to the distances of
the corresponding points on the V.T. from XY ; e.g. to obtain
the point C, set off /tC equal to e'c\ and so on with each of the
other points. Join them, as shown, to complete the construction l
of the sectional surface.
To rotate the sectional surface on the V.T. Draw the perpen
diculars from the points e\ a\ ^, /', and make the lengths of
these lines equal to the distances of the corresponding points
below XY ; e.g. make/'G' equal to mg.,f'Y' equal to inf., and so
on with the other points. Join them as shown.
220. To project a quadrilateral prism 10 V' long, with a base 6"
square, standing on its base on the H.P., with its longer edges
parallel to the v. p., and one of its sides inclined to the v. p. at
an angle of 60^ Intersect the prism with a plane parallel to
XY, and inclined to
the H.P. at an angle
of 30'. Scale J^ full
size. Fig. 302.
Project the prism
(Prob. 171). Find the
traces of the cutting
plane (Prob. 218).
At any convenient
point e draw the line
E/ at right angles to
XY cutting the H. and
V. traces. With e as
centre, and radius ^E,
draw an arc till it meets
XY vcif. Join e'f. efe'
Fig. 302.
is the angle the cutting plane forms with the H.P.
Draw lines from each point in the plan perpendicular to ^E,
and meeting it in the points C, B, D, A. With e as centre, and
each of these points as radii, draw arcs ; and where they meet XY
draw perpendiculars till they meet ^yin the points C', B', D', A'.
SECTIONS OF SOLIDS.
171
From these points draw horizontal hnes till they meet projectors
drawn from the corresponding points in the plan in the points
c\ b\ d^ a'. Join these points for the elevation of the section.
To construct the sectional area. Draw perpendiculars to e'/aX
the points C, B', D', A', and make their lengths equal to the
distances of the corresponding points in the plan from the line
^E ; e.g. make C'C" equal to Cf, B'B" equal to B<^, and so on
with the other points, and join them.
221. Project a regular hexagonal prism 9V' long, with Stt" sides,
standing on its base on the H.P., with one of its faces inclined
to the V.P. at an angle of 58 . Intersect the prism by a plane
inclined at an angle of 55' with the H.P., and 46 with the v.p.,
the plane to cut through the base of the prism. Construct on
the H.P. the sectional area. Scale ^.i full size. Fig. 303.
Project the prism (Prob. 185). P^ind traces of cutting plane.
Draw lines from the points of the plan parallel to the H.T.
till they meet XY, then draw perpendiculars to XY, till they
meet the V.T. in the points
C, A', D', F', E'. From
these points draw horizontal
lines till they meet projectors
drawn from the correspond
ing points of the plan in the
points c\ d\ e\f\ a\ b\ and
join them as shown. This is
the elevation of the section.
To construct the true shape
of the section, we rotate the
plan abcdef on the H.T. From
e draw the line eg parallel to
the H.T., and equal in length
to the height of ^'above XY.
Draw lines from each of
the points in the plan at right
angles to the H.T. : the one
drawn from e will intersect the H.T. in //. With this point as
centre and the radius kg., draw an arc till it meets the line from
e produced in E. Find all the other points in the same way as
shown.
Fig. 303.
172
GEOMETRICAL DRAWING AND DESIGN.
The student should observe that the Hue JiK is the true
length of the line he. A simpler method of obtaining the
length of these lines is as follows.
Note. — The true length of a projected Hne is equal to the
hypotenuse of a rightangled triangle, the base of which is
one of its projections, and the altitude the perpendicular height
of the other projection.
Let kd represent the horizontal projection of a line, and Yid'
its perpendicular height. From k set off on the H.T. k7J equal
to Kd'. Then the distance between 7i and d will be the true
length of the line kd. Set this distance off from k on dk
produced. This will give the point D. The other points
can be found in the same manner.
222. Project a regular hexagonal prism 10 V' long, with 3^
lying on one of its faces on the H.P., with its longer edges
inclined to the v.p. at an angle of 17°. Intersect the prism
with a plane inclined at an angle of 50° to the h.p. and 56° to
the v.p. ; the plane to
intersect all its longer
edges. Construct on the
v.p. its sectional area.
Scale jV full size. Fig.
304
Project the prism (Prob.
186).
Find the traces of the
cutting plane (Prob. 218).
Let BC be the side of the
prism resting on the H.P.
Join AD. Then BC, AD,
and EF are horizontal lines,
and B'A' and B'F' their
heights above XY.
Note. — A horizontal Hne
contained by a plane would
Fig. 304 be drawn parallel to the H.T.
on plan, and parallel to XY in elevation.
As be is on the H.P., it must coincide with the H.T. Where
the lines F' and A' produced meet the V.T., drop perpendiculars
SECTIONS OF SOLIDS. 173
till they meet XY, and then draw lines parallel to the H.T.
Where these lines intersect the lines of the plan they will
determine the points of the section. Join them as shown.
Carry up projectors from these points till they meet the
corresponding lines in the elevation, and join them.
Note. — The projectors are omitted in several of these
problems to save confusion, but the points in plan and eleva
tion bear corresponding letters throughout, so they can be easily
recognised.
To construct the sectional area on the V.P. Draw lines from
each of the points in the elevation at right angles to the v.T.
Take the distance of point e below XY as ^E', and set it off on
the v.T. from g as gh. Set off gY!' equal to he\ the hypotenuse
of a rightangled triangle, as described in the preceding problem.
Find the other points in the same way, and join them as shown.
223. Project a regular pjrramid standing on its base on the H.P.,
with its sides inclined to the V.P. at an angle of 45' ; the cutting
plane to be inclined at an angle of 43^ to the H.P. and 70° to
the V.P. Construct the sectional area on the V.P. Fig. 305.
Project the pyramid (Fig. 248).
Find the traces of the cutting plane (Prob. 218).
Produce the diagonal eg till it meets the H.T. in E. Draw
the projector EE', and from E' draw a line parallel to the V.T.
Where this line meets the edges of the pyramid in b' and d' will
determine two points in the section. Drop projectors from
these points till they meet the corresponding hues of the plan
in the points b and d.
The lines forming the section of this pyramid are really the
intersection lines of two planes ; the cutting plane being one,
and each side of the pyramid the other plane. We know that
the line of intersection between two planes must have its .trace
where the traces of the two planes intersect. Produce three
sides of the base of the pyramid till they meet the H.T. in
the points k^ m, and /. These are the traces of the lines
required. Draw a line from k through b^ and produce it till
it meets the diagonal fh in a. Draw lines from jn and / in
the same manner till they meet the diagonals in d and c.
Join dc. This will complete the plan of the section.
174
GEOMETRICAL DRAWING AND DESIGN.
To determine the elevation of the points a and c. With
centre o^ and radii oa and oc, draw arcs till they meet the
diagonal es[ in the points 7t and/. Draw projectors to these
points till they meet the edges of the pyramid in the points n'
Fig. 305.
and p'. Draw horizontal lines from these points till they meet
the other edges of the pyramid in the points c' and a!. Join
the points as shown, to complete the elevation of the section.
Another method of obtaining the section of this pyramid is
to assume a horizontal line q'r' in any convenient position in
the elevation, and drop a projector from r' till it meets the
SECTIONS OF SOLIDS.
175
diagonal eg in r. Draw rt parallel to the base gh. Produce
q'r' till it meets the v.T. in /. Draw a projector from / till it
meets XY in s. Draw a line from s parallel to the H.T. till it
meets the line r/ in 11. Then ii is a point in the plan of the
section, which can be completed from the traces k, ;;/, and /, as
previously described.
To construct the sectional area ABCD, proceed in the manner
described in the preceding problem.
224. To project a section through a right vertical cone : the cutting
plane to be perpendicular to the v. P. ; to be inclined at an
angle of 45° to the axis of the cone, but not to intersect its
base. Fig. 306. This section is an ellipse.
Let DE be the elevation of the section. Divide it into any
Draw the axis of the cone,
'%
•if
number of equal parts, e.g. six.
and through the divisions on DE
draw lines parallel to the base of
the cone.
The plan of the section is
determined by first finding a
succession of points in the curve,
and then drawing a fair curve
through them. We will take the
points dd as an example. With
C as centre, and a radius equal
to G'H {/.e. the radius of the cone
at the level of &'), draw an arc till
it meets a projector drawn from
d' in the points dd. Proceed in
the same manner with the other
points, and draw a fair curve
through them. Fig. 306.
To construct the sectional area. Draw the line D'E' in any
convenient position, parallel to DE, and draw hnes from each
of the divisions on DE at right angles to D'E'. Take the
distance Gd from plan, and set it off on each side of G" in the
points d"d". Find all the other points in the same manner, and
draw a fair curve through them.
176
GEOMETRICAL DRAWING AND DESIGN.
225. To project a section through a right vertical cone ; the cutting
plane to be parallel to the side of the cone and perpendicular
to the V.P. Fig. 307. This section is a parabola.
Let D'E' be the elevation of the section. Divide it into any
number of parts — it is better
to have the divisions closer
together towards the top.
Draw horizontal lines through
these divisions.
The plan is determined by
finding a succession of points
as in the preceding problem.
We will take the points bb
as an example.
With C as centre, and a
radius equal to the semi
diameter of the cone at the
level of the division b\ i.e.
G'H, draw an arc till it meets
a projector from b' in the
points bb. Find the other points in the same manner, and draw
a fair curve through them.
To construct the sectional area. Draw the hne D"E" in any
convenient position parallel to D'E', and draw lines at right
angles to it from the divisions on D'E'. Take the distance
Qb from plan, and set it off on each side of G" in the points b"b".
Proceed in the same manner with all the other points, and draw
a fair curve through them.
Fig. 307.
226. To project a section through a right vertical cone; the cutting
plane to be perpendicular to the H.P, and inclined at an angle
of 50° to the V.P. Fig. 308. This section is an hyperbola.
Let DE be the plan of the section. From C draw the line
(Zd perpendicular to DE. With C as centre, and radius Qd^
draw an arc cutting AB in n. Draw the projector nn' . n' is
the vertex of the section. Divide the height g'ji' into any
number of divisions, — they should be made closer together near
the vertex, — and draw horizontal lines through them till they
meet the sides of the cone.
SECTIONS OF SOLIDS.
77
The elevation of the section is determined by first finding a
succession of points, and then drawing a fair curve through
them. We will take the points b'b' as an example.
With the point C on plan as centre, and a radius equal to
G'H (the semidiameter of the cone at the level of b'\ draw arcs
intersecting the line DE in the points bb. Draw projectors to
these points till they meet the line drawn through G'H in the
points b'b' . Find the other points in the same manner, and
draw a fair curve through them.
If the cutting plane were perpendicular to both the coordinate
planes, gh would be the plan, and g'ji' the elevation of the
section of the cone.
To construct the sectional area. Draw the line ii'g" in any
convenient position parallel to the axis of the cone. Produce
178
GEOMETRICAL DRAWING AND DESIGN.
the divisions on 7ig' . Take the distance db from plan, and set
it off on each side of G" in the points b"b". Find all the other
points in the same way, and draw a fair curve through them.
As the three preceding problems are conic sections, their
sectional areas could be constructed by the methods described
in Chap. XI. (Plane Geometry), but we must first determine
the major and minor axes of the ellipse, and the directrices and
foci of the parabola and hyperbola.
We will illustrate by a perspective view (Fig. 309) the
principle of the relation between the directrix and focus of a
parabola, and after
wards apply it to the
ellipse and hyperbola.
ACB is a cone, and
DEGF the cutting plane.
H is a sphere touching
the cutting plane, and
inscribed in the upper
portion of the cone. A
line drawn from c, the
centre of the sphere,
perpendicular to the
cutting plane, will meet
it in /, which is the
focus of the parabola.
The plane KLNM,"
containing the circle of contact between the sphere and the
cone, intersects the cutting plane in the line ag^ which .is the
directrix of the parabola.
A line joining the centre of the sphere with the circle of
contact, as ce^ is perpendicular to the side of the cone.
Compare this figure with Fig. 228 (Plane Geometry).
Let us now refer to Fig. 307. cfe is the inscribed sphere, /is
the point of contact with the cutting plane, ce is perpendicular
to the side of the cone, and e determines the level of the plane
containing the circle of contact. A horizontal line drawn
through e till it meets the cutting plane produced in d will
determine the position of the directrix.
Draw a line from y" perpendicular to the cutting plane till it
SECTIONS OF SOLIDS.
179
meets the line D'E' in /' : this is one of the foci of the elhpse,
A hne drawn from d perpendicular to the cutting plane will
determine the directrix ag.
The line D'E' is the major axis of the ellipse, and if we
bisect this line by another at right angles to it, and obtain the
position of the points K and L in the same manner as we
determined the points b"b'\ KL will be the minor axis. We can
obtain the other focus and directrix by setting off their distances
on the opposite side of KL ; or we could construct another
sphere in the lower part of the cone, and obtain them as
already described.
In Fig. y:>'j^ the same construction as previously described
will determine the position of the directrix and focus ; and as it
bears corresponding letters, the student should have no diffi
culty in understanding it. Compare Fig. 307 with Fig. 309.
The same thing applies to Fig. 308.
227. To project the section of a riglit vertical cylinder ; the cutting
plane to be inclined at angle of 36' with the H.P. and 73' with
the V.P., hut not intersecting the base. This section is an
ellipse. Construct the sectional area on the H.P. Fig. 310.
Project the cylinder
(Prob. 203).
Find the traces of
the cutting plane (Prob.
218).
Assume a vertical
plane passing through
the axis of the cylinder,
and perpendicular to the
cutting plane. Draw ab
perpendicular to the H.T.
Draw a projector to a till
it meets the v.T. in a .
Draw a projector to b till
it meets XY in b' . Join
cib'. ab is the H.T. of this
V. plane, and ctb' the line in which it intersects the cutting plane.
Draw a projector to o till it meets ab' in o' .
i8o GEOMETRICAL DRAWING AND DESIGN.
Divide the plan by diameters into eight equal parts, one
of these diameters, dh, being in the H.T. of the V. plane.
Projectors to dh will give d\ h', two of the points in the section.
Produce the diagonal ie till it meets the H.T. of the cutting
plane in 7n. Draw the projector in7n\ and draw a line from in'
through d. This will give the corresponding points z\ e\ in
the elevation.
Produce the diameter jf till it meets XY in fi. Draw a
perpendicular to XY at ft till it meets the V.T. in n'. Draw a
line from Jt' through d. This will give the corresponding points
/',/' in the elevation. Find the points c'^g* in the same manner.
Draw a fair curve through these points for the elevation
of the section.
Any number of points in the curve could be found in the
same manner by drawing additional diameters to the plan,
but eight points are generally deemed sufficient.
To construct the sectional area. Find the points B, H, F, D,
(Prob. 221), and complete the ellipse (Prob. 1 57, Plane Geometry).
228. To project the section of a sphere ; the cutting plane to be
inclined at an angle of 35' with the H.P., and 74' with the V.P.
Fig. 311.
Find the traces of the cutting plane (Prob. 218).
As a sphere is a continuous surface without any edges or
angles, it will be necessary to assume certain fixed lines upon its
surface in order to determine where the cutting plane will inter
sect it ; meridians and parallels are best suited for this purpose.
Project the sphere with meridians and parallels (Prob. 207).
It will be better to arrange the meridians on plan so that one
of them is parallel to the H.T.
Assume a V.P. perpendicular to the cutting plane and con
taining the axis of a sphere. Let ab be the H.T. of this plane,
and db' the line in which it intersects the cutting plane. Where
this line intersects the axis will determine <?', and where it
intersects the meridian dd' will give two points in the section.
Produce the meridian jk on plan till it meets the H.T. in ?t.
Draw the projector im', and draw a line from 7t\ through o\
till it meets the meridian j'k'. These are two more points in
the section.
SECTIONS OF SOLIDS.
iSi
Produce the meridian ef till it meets XY in /. Draw a
perpendicular at / till it meets the V.T. in /'. Draw a line
from /', through o\ till it meets the meridian f e\ giving two
more points in the section. Obtain the points h' and g' in the
same manner, and draw a fair curve through all the points found.
Fig. 3".
Drop projectors from each of these points till they meet the
corresponding meridians on plan, and draw a fair curve through
them.
These projections are ellipses. They could also be found by
first projecting their conjugate diameters, and then completing
them as in Problem 157.
The true shape of the section is of course a circle. To obtain
its radius, bisect cd in j, and draw a line tii through s parallel
to the H.T. ; st is the radius required ; tu is the major axis, and
/r^the minor axis of the ellipse.
I82
GEOMETRICAL DRAWING AND DESIGN.
EXERCISES.
1. Draw the plan of a right pyramid 2j" high, base an equilateral
triangle of 2" side. The pyramid stands on its base, and the upper part
is cut off by a horizontal plane i" above the base. Indicate the section
clearly by light shading. (April, '96.)
2. An elevation is given of a regular hexagonal prism with its bases
horizontal (Fig 312). Draw its plan. Also show the true form of the
section made by the inclined plane shown at bn. (April, '00.)
Fig. 312
Fig. 314
3. The diagram (Fig. 313) shows the end elevation of a right
prism if" long with square base, and a horizontal plane Im cutting the
prism. Draw a plan of the portion below Im. (April, '98. )
4. The diagram shows (Fig. 314) the plan of a portion of a sphere.
Draw an elevation upon the given line xy. (June, '98.)
5. Plan and sectional elevation are given
(Fig. 315) of a short length of moulding which
has been cut across for " mitreing " by a vertical
plane shown in plan at ////. Determine the true
form of the section. (June, '97.)
^ 6. A right square pyramid, edge of base i",
height 2", is cut by a plane which contains one
edge of the base, and is inclined at 45° to the
plane of the base. Draw the plan of the section,
and if you can its true form. (June, '99. )
7. A sphere of \" radius has a portion cut off
by a horizontal plane g" above its centre, and
another portion by a vertical plane passing ^"
from the centre. Draw a plan of what remains
Fig. 315 of the sphere. The section must be clearly
indicated by lightly shading it. (June, '97. )
8. A right cone, height 2^" , radius of base i", stands with its base on
the horizontal plane. A sphere of i" radius rests on the horizontal
plane and touches the cone. Draw a plan and elevation of the two
solids, showing clearly your construction for finding the centre of the
sphere. Show also the true form of the section of each solid by a
horizontal plane ^" above the centre of the sphere. (April, '02.)
i^For fmther Exercises see p. 228. )
PART III.
DESIGN.
CHAPTER XVI.
The eye and the ear are both pleasantly affected by regularity
of effect. The ear treats as music the even beats of the air, and
in the same way the eye is a very accurate judge of evenness
and regularity. It is strange how unevenness distresses the eye ;
for instance if the hnes of this page were unequal distances
apart, the effect would be very irritating.
But there is a further parallel between music and de
coration. A discord has an effect on the enr which may be
described as unsatisfying, and the "Resolution of the Discord"
is called for to satisfy the demand for evenness, which has
been temporarily violated.
In the same way, in decoration, the eye demands regularity,
even uniformity of treatment ; while at the same time it has a
feeling that unevenness or the interruption of uniformity is
pleasurable, if supplemented by regularity in the whole design.
The two precisely similar spires of Cologne or Coutances are
not altogether satisfying, because of their want of contrast ;
the eye is not challenged by contrast into attention to detail.
This constitutes the special beauty of the twin, though dis
;similar, spires of Lichfield or St. L6, where the eye is piqued by
the discord into the discovery of the prevailing harmony of
■design.
These principles should guide us in decorative design.
Regularity must prevail in the main, with the unexpected to
afford and challenge interest.
i84 GEOMETRICAL DRAWING AND DESIGN.
Construction Lines on "w^hich Patterns are
arranged.
To cover a wide space with regularity of form is an ancient
device of man ; and in fact the old Greeks were better versed
in its secrets than are we of this age. And yet, if we note
their methods carefully, we may observe that geometrical
balance was the secret of their success. To map out a sur
face with a geometrical network, and to fill it with balanced
forms satisfying to the eye, seemed to them second nature.
Slowly we are regaining the same instinct, and the beautiful
decorations of modern artists are the product of dihgent study
of geometrical design.
1
1
Fig. 316.
Fig. 317.
Fig. 318.
Fig. 319.
A net is the simplest aid to mapping out a surface, and we
have in Fig. 316 this groundwork of design, a square net like
a tennis net ; and in Fig. 317 a 45° net, rectangular, and each
hne making equal angles of 45° with the horizon. Fig. 318 is a
60° net, covering the surface with a vertical diamond or lozenge :
Fig. 319 a 30° net, forming a horizontal diamond or lozenge.
This mapping out of the whole surface by a network, or skeleton,
must be the first step, as it is the geometric basis for the
pattern, the unit of which will be the subject of discussion later.
DESIGN.
i8s
The geometric framework must be prepared before the pattern
which it is to carry. It will be noticed that in these nets certain
lines are emphasized, showing how the simple net may be the
basis of a more complicated network. A square net with
xxix
><
M>^>^jx
y
><i
X!xp^
Vy/
W\y\;^Ay'
\^
V
XP\D\
Y
vvv
>sp\5s
^K'Os
XXx
>$><p<.
X
44^
>^A>\
X
xtxp!^.l
xxx
X*
XN\
^i^x
XiXX
X
xlxx
xlxlx
Fig. 320.
Fig. 321.
Fig. 322.
Fig. 323. Fig. 324.
1
1
1
■ jrn.

1 1
n
1 1
n
 i_
Fig. 326.
Fig. 327.
diagonals is shown in Fig. 320, and m Fig.
^,21 a
45 net
crossed by vertical and horizontal lines, which form octagons
and other figures In both Figs. 320 and 321 a square net and
a 45° net are combined and certain hues are selected. In Fig.
322, a 60° net with horizontals forms a network of equilateral
triangles or hexagons. Fig. 323 is a square lattice, and Fig.
324 a square plaid ; Fig. 325 a lattice of 45° ; Fig. 326 a double
1 86
GEOMETRICAL DRAWING AND DESIGN.
square lattice, and Fig. 327 a square framework. In these we
see how the choice of Hnes in a square net fills a whole area
with a skeleton, a process which is simply exemplified in
Fig. 328.
X
m
X
m
X
m
Fig. 329.
1/
\
xxxx
A
X
/
\\,
X
A.
X
X
^
/;
X
X
/
^
X
7
^
\
s
XXX
\
/
9>.
\
\^
/
/
4
\
/
K TV
Fig. 330.
Fig. 331.
Fig. 332.
Fig. 333
Fig. 328. Fig. 329 shows a skeleton formed by a square and
diagonals, forming interlacing octagons ; Fig. 330, a roof tiling
made by the choice of diagonals on a square net ; Fig. 331 a
diaper constructed in the same way. Fig. 332 is a triangular
DESIGN.
187
framework formed on a 60° net ; the upper part is a drop
triangular framework, the triangles are only blackened to define
the triangles of the design ; the lower part shows a series of
interlacing triangles, identified by their blackened centres.
Fig. 333 is a hexagonal framework on a 60^ net, with verticals.
There are three small hexagons to each triangular form. The
network, shown in the lower righthand corner, being omitted,
the lines left form an elaborated skeleton of lines, chosen
regularly and uniformly related to one another.
^
Fig 335
^ Z S Z \
/ \ / \ /
1
S zs_z s
V i;/' ^ ^
1 11
000
moo
Fig. 337
Fig. 334 is a framework of interlacing triangles forming a
star and outlining a hexagon. Fig. 335 shows a framework
in which interlaced stars all fill up the pattern and give a
hexagon form ; Fig. 336, a framework of unequal sided octagons,
formed on a square net. In Fig. 337 a framework of squares
and stars is formed on a square net which is shown in one of
them ; the diagonals of the net construct the stars, part only of
the diagonals being shown. Figs. 338, 339 and 340 show the
development of simple line patterns, suitable for band ornament,
formed on a network. In Fig. 341 the selection of lines shows
i88
GEOMETRICAL DRAWING AND DESIGN.
the development of a pattern with the network omitted, and
Figs. 342, 343 and 344, the selection of the rectangular frame
work and added diagonals. Fig. 345 is a plaited band on a
60° net.
Fig. 338.
Fig. 330.
MM"i"n
IIIIII'MI —
Fig. 340.
Fig. 341 Fig. 342.
/__y:
Fig. 344
Fig. 343
AAAAAAA7
Fig. 345.
Thus far the skeleton or framework has been composed of
lines selected from a network or diagonals of the net, and formed
of straight lines only. It remains to add the circle to the
materials of construction and to proceed to form frameworks
with its aid.
Fig. 346. Semicircles are worked on the square net giving
scale work or imbrication, and Fig. 347 shows a scale work on
a 30° net.
DESIGN.
189
Fig. 348 shows a framework on a 45° net formed of interlacing
circles.
Fig. 349 suggests how, on a 60° network with horizontals, a
pattern can be developed from interlacing circles and straight
lines outlining a hexagon.
^^
sz
"O"
^^
^p
K
^^^^
^Xm
»
^
\
Fig. 351
Fig. 349
Fig. 352.
Fig. 35
Fig. 356.
Figs. 350353 give frameworks of circles on a square net
the net of construction being indicated by dotted lines.
Fig. 354 shows the change which takes place in the framework
by working with circles on a 45° net.
Fig. 355 is a skeleton developed from selected semicircles on
a square net, and Fig. 356 the same skeleton with additional
semicircles which cover the vacant intervals.
I go
GEOMETRICAL DRAWING AND DESIGN.
Fig. 357 is a complete skeleton ot interlacing circles and is a
pure circular pattern. This pattern is based on a 60° net, and
it is an example of a simple pattern which a child with a pair
of compasses could draw as he fills a paper with circles.
Fig. 358.
^:
X^
J
V
J
x^x
^
\
^
y
J
V
J\^^ )
^
C
~\
r ^^^''"'^
Fig. 359
Fig. 360.
Fig. 358 is the result of selecting interlacing circles based on
the 30° net and forms a beautiful geometrical pattern.
Fig. 359 is a scale work on a square net, the radii of the circles
being less than the side of the square.
Fig. 360 is an ogee skeleton and Fig. 361 part of a framework,
both resulting from the selection of semicircles on a square
net.
In Figs. 362 and 363 the selection of quadrants on a square
net construct two very different skeletons of interlacing ogee
forms.
DESTGN.
Fig. 364 is a skeleton composed of circles and semicircles on
a square net.
Figs. 365367 show an effect of combining circles and straight
7
5<<
^ ^ ^
QXy^
^ 2^
±
Fig. 361.
Fig. 362.
^j^^j^^j^
¥
TTT^
Fig. 363.
Fig. 364.
Fig. 365
Fig. 366.
Fig. 367
lines on a square net ; the net being omitted, a framework or
skeleton remains to receive the details of ornament.
Figs. 368371 show the effect of working on a square net, with
circles, semicircles and straight lines.
192
GEOMETRICAL DRAWING A.ND DESIGN.
Fig. 372 is a double scale work formed by concentric semi
circles on a square net.
Fig. 368.
i»
/
A
V \r
Fig. 369.
<
^
V
xx>c
^
A
Y
y
v^
/^
^
X
^1
V
V
r
N
r
^
—
Fig. 370.
Fig. 371
1
^
X?ft
L
^^
^
J
"^Yr y/r ^^
r
■N^
.yrvL. jYv^ yr
^
^
1
^A
^r V:^ v^
372.
F.g. 373
Fig. 373 is a framework or skeleton border on a square
net.
Fig. 374 IS a framework of quadrants and semicircles with
diagonals on a square net.
DESIGN.
193
Fig. 375 is a simple roll border formed by parallel lines and
concentric circles on a 45° net.
Fig. 374
Fig. 375.
Fig. 376.
Fig. 377
Fig. 378.
Fig. 379.
Fig. 380.
Fig. 381.
Units of Pattern.
Having shown how to map out a surface by a network or
skeleton, in varied ways, we must now proceed to discuss the
units of pattern with which the surface is to be covered. Figs.
376, 377 and 378 are simple circular designs, the construction of
which is obvious ; in Fig. 376, the circles are joined by freehand,
forming a ballflower ornament which occurs frequently in
' Decorated ' Gothic architecture. Fig. 379 illustrates the varia
tion of pattern caused by emphasising different parts of a
design ; the two halves of this unit are really different patterns,
though the lines in both are the same. Fig. 380 is a triangular
194
GEOMETRICAL DRAWING AND DESIGN.
unit ; for its construction refer to Problem 'j']^ page 49, a triangle
drawn to touch the dotted circle at CEG gives
the outline. Fig. 381 is an ogee unit constructed
on a square and its diagonals. Fig. 382 is a
square unit, and Figs. 383, 384, units of inter
secting squares. Fig. 385 is a unit constructed
z L^^z:
Fig. 383.
Fig. 38A.
Fig. 386.
Fig. 387.
Fig. 391.
Fig. 389
on a 3 by 2 rectangle ; Figs. 386, 387, 388 are square
units formed by a combination of circles and straight lines.
Fig. 389 is a square lozenge, enclosing intersecting circles ;
Fig. 390 consists of concentric squares and circles, all enclosed
in an equilateral octagon. Fig. 391 is a star formed by two
interlacing equilateral triangles, with parallel but not equidistant
DESIGN.
'95
sides ; Fig. 392 is a square star unit ; Fig. 393, a unit of interlacing
squares, enclosing a circular design. Fig. 394 is a star hexagon
unit. In Fig. 395 geometrical design and freehand are combined
on an octagon base.
Fig. 393
Fig. 394
Fig. 395
These few examples of geometrical units of pattern must
suffice to show the principles of construction. Naturally enough,
the scheme of decorative treatment usually leaves the trammels
of geometrical design, relying upon the forms of flowers and
foliage for the ideas which the repeating patterns carry out.
But the object in view has been attained if the student has been
led to see, under the intricacies of decoration, the geometrical
basis on which it is constructed.
{Fo7 Exercises see p. 229.)
196
GEOMETRICAL DRAWING AND DESIGN.
The Spacing of Walls and other Surfaces.
The wall of a modern dwellinghouse is usually divided, for
decorative purposes, into the cornice, frieze, field, dado and
skirting, as shown in Fig. 396.
Skirting
Fig. 396.
The various methods for covering such given spaces with
ornamentation by means of geometrical patterns are briefly
indicated in this section. For the technique of distributing and
repeating patterns, suitable for such special spaces, in any
decorative scheme, the student should consult any of the various
books on Decorative Design.
DESIGN. 197
This chapter aims at teaching him how to construct patterns
geometrically : it is quite another and a larger subject which
must be separately studied how to make geometrical treatment
subservient to the decorator's art.
tmnnj 'gfHTEl nSZJ
Fig. 397. Fig. 398. Fig. 399
mm.
Fig. 400. Fig. 401. Fig. 402.
Fig. 403. Fig. 404.
JjOOQinki
Fig. 405. Fig. 406.
Fig. 407.
(1). Bands and Borders.
A wall is very often divided into two parts by a horizontal
border above the dado ; the treatment of such a border is a simple
introduction to wall decoration. We begin with a set of bands
called "Greek Frets"; these are represented in Figs. 397 to 405.
They are formed by selecting lines from a square net ; an
introduction was made to this in Figs. 338, 339, 340, where the
lines of construction are seen. In Fig. 401, a raking pattern, the
198 GEOMETRICAL DRAWING AND DESIGN.
diagonals of the net are utilized for construction. Figs. 406408
are intersecting frets, 406 being a Moorish plaited band and 407
an Italian interlacement band ; 408 is a chain band, and 409 a
^ [7
^
U
£}
Fig. 40S.
ig. 409.
Fig. 418.
straight line band suggesting a series of Maltese crosses formed
of onyxes.
With Fig. 410 a series of circular band ornaments is intro
duced, leading up to the waveform introduced in Fig. 414.
DESIGN,
199
The wave and roll form is developed in Figs. 415 to 418,
and its construction carried on progressively to the complicated
rolls, Figs. 419, 420. Fig. 421 shows one of the effects of a
Fig. 419.
Fig. 420.
Fig. 421,
double roll combination suggesting to the student what a
variety of interesting designs may be evolved from the con
structions thus built up. The term "Guilloche" is generally
apphed to such rolls as are shown in Figs. 419421.
200 GEOMETRICAL DRAWING AND DESIGN.
With Figs. 422424 is commenced a series of bands having for
foundation the combination of straight and curved lines, leading
t>^^ I iF^^^rx
Fig. 422.
Fig. 423.
Ml 10
Fig. 424.
Fig. 425.
Fig. 426.
i::z:
1^
Fig. 427.
Fig.
Fig. 429.
^_
Fig. 430.
V.^
■^ X
Fig. 431
Fig 434
^p»n^ ^
Fig. 432.
Fig. 433
up to Figs. 425, 426, in which the effect of such combinations
in decoration is exemplified.
DESIGN.
Spiral elements are introduced in Figs. 427 to 430, and the
wave line in Fig. 431 ; and with these materials the beautiful
bands given in Figs. 432 to 439 are built up. Figs. 432, 433
and 434 are Greek paintings on terracotta. Fig. 435 is a
I'"!^' 435
Fig. 436.
Fig. 437
Fig. 43S.
Fig. 439
French mural painting of the 13th century; Fig. 436, a border
from a picture by Domenicino (i6th century). Fig. 437 is a
Greek terracotta, and 438 an Early Gothic French ornament.
Fig. 439 is an "aesthetic" design. The two "repeats" in these
cases are a sufficient guide to the complete scheme.
{For Exeirises sec p. 231.)
202 GEOMETRICAL DRAWING AND DESIGN.
(2). Defined Areas— Walls, Cerings, Floors, etc.
We come now to deal with the treatment of certain defined
areas ; hitherto we have only dealt with schemes meant for
general areas. Fig. 440 indicates the simplest division of a
square so as to give border and corners, while Fig. 441 only
Fig. 440.
Fig. 441.
Fig. 442.
Fig. 443
Fig. 444.
Fig. 445
Fig. 446.
affords a plain border, inside which a Maltese cross divides up
the space for further ornamentation. Fig. 442 shows a circular
centre, and Fig. 443, circular corners. Fig. 444 is a simple indi
cation of the centre and corners due to the inscribed circle, while
Fig. 445 shows an elaboration of the centre with wide borders.
Fig. 446 is a square design, with a centre which arrests the
DESIGN.
203
eye by being unexpected, as the arms are not radial. Fig. 447
is a square ceiling design ; the top outer circle is completed to
show the construction, the true design being given in the lower
part. Fig. 448 is a panelling for a ceiling from a tomb in Rome,
forming, as in all the cases we have been considering, a skeleton
n ^ /
DUOL_
D
D
Fig. 447.
Fig. 448.
Fig. 449.
Fig. 450.
Fig. 451.
XX
Fig. 452.
i_j
\2^^n
Fig 453
for decorative treatment. Fig. 449 is a rectangular space which
is divided up by selected lines of a square net, with certain
diagonals. Fig. 450 is a square lozenge, separating the corners
of the rectangle. Fig. 451 is a cross of St. Andrew, with
circular centre, affording triangular panels. Fig. 452 shows a
rectangular moulding with subdivision of the space in the
204
GEOMETRICAL DRAWING AND DESIGN.
form of a cross. Fig. 453 shows a simple circular treatment
of a panel ; Figs. 454, 455, two lunettes with circular treatment.
Fig. 456 is a simple lunette and spandrels. Fig, 457 gives a
subdivision of a circular space suggesting tracery, and Fig.
458 a trefoil treatment. Fig. 459 gives the complete hexagonal
system of circles, inscribed in the given area, from which the
artist may select or emphasize symmetric arcs, so as to produce
Fig. 454
Fig. 455
Fig. 456.
Fig. 457
Fig. 459
Fig. 460.
an extraordinary variety of design. Fig. 460 is a tracery design,
worked from the hexagon of the outer circle. Fig. 461 is the
subdivision of an octagonal area, the figure being built on a
square net ; Fig. 462 on the 45° net. Fig. 463 is a star figure,
which is built up inside an octagon. This is a sample of the
many dififerent stars which may be formed by varying the radius
of the dotted circle of construction. The student will find a
DESIGN
205
useful exercise in making several examples, which will illustrate
the various designs which result from the change of this circle.
For example, if the radius of the dotted circle be made about
half of that in Fig. 463, and radii of the larger circle be drawn
to the points of the star, we have the familiar appearance of the
old mariner's compass as it used to be before the advent of
the spider web of the Thomson compass.
Fig. 46
Fig. 462.
Fig. 463.
Fig. 464.
Fig. 466.
Fig. 464 again is a sample of the division of a hexagonal
space which is suggestive of many varieties. The elementary
feature, the joining of all the points of a hexagon, suggests in
appearance the outlines of a transparent icosahedron : a star,
inscribed in a circle is formed in the hexagon. The figure itself
can be amplified by further outlining the details, the simple
plan adopted in the figure is to follow each line with another
line parallel to it at a fixed distance throughout. It is evidently
possible to produce a large variety of divisions of the hexagon
on this model, and the introduction of circular arcs will add still
2o6 GEOMETRICAL DRAWING AND 3i::
more. L 5, 466 the space treated is an equilateral
triangle ; here a. simple plan is to divide it either by a central
hexagon, as in Fig. 465, or by a central circle, as in Fig. 466.
In each of the examples, th« feature of doubled and parallel lines
is introduced.
The object which is airied at in presenting these figures is
to suggest modes of setting out defined areas — walls, ceilings,
and floors — for decorative treatment. It is this general
delineation which is the particular aim of geometrical design.
In the completed scheme, no doubt, the framework or scaffolding
will be entirely lost, but the aim of the designer is to afford a
pleasing arrangement of the space available ; and in such a
way that the eye should not be arrested by the ornamentation
forming a simple network.
It has often been observed that a wall paper for a sick room
must not offend in this particular. If a patient's eye is con
tinually challenged by a repeat in network over the surface, or
a figure pattern too prominently recurrent over a large area, the
mental disturbance is serious. The mind falls to counting the
patterns in wea:risome persistence, and the more the mind is
beyond physical control, the more serious is the effect.
Hence it is desirable that the designer should practise the
division of spaces by forms which have a pleasing intricacy of
geometrical balance. The methods here indicated are only
specimens which should suggest the lines which may be
followed, and it is hoped that they will lead the student to
exercise his ingenuity in planning more serious designs.
[^For Exercises see p. 232.)
TION.
ORNAMENTATION.
CHAPTER XVII.
Lettering.
It is very important for the student to be able to letter his
drawings well. The design of Roman lettering is a serious
study of which it is possible only to give the merest outline
in Figs. 467, 468. The basis of construction is a square of
TX
^
Fig. 467.
Fig. 468.
rectangular net, with circles to guide in the formation of the
serif The revival of the old style of lettering, with the serif
inclined as shown in Fig. 467, letter L, has given a new and
interesting impetus to the artistic study of lettering.
These two examples are taken from Albert Durer's Geometrica^
in which he gives methods for drawing Roman capitals.
Suppose as a groundwork a square. The thick strokes
are \ of the square and the thin strokes are ^^ of the
square. The serifs are constructed on circles of j
diameter. Mr. Walter Crane says : " Letters may be ...ken
as the simplest form of definition by means of line. They
have been reduced through centuries of use from their pi' nillve
hieroglyphic forms to their present arbitrary and fixed vpes :
though even these fixed types are subject to the Vitriation
produced by changes in taste and fancy."
2o8 GEO "LTRICAl^' DRAWING AND imSOJ^.
Shields.
Shields often appear as an element of ornamentation, and
must be treated in accordance with the rules of heraldry. It is
a very early and general rule that metal must not be placed
upon metal, nor colour upon colour ; but that they must be
placed in contrast. ' Or,' gold, and ' argent^ silver, are the
metals used, and ' aziire^ blue ; gules^ red ; ^ purpurea ' vert.^
and ' sable ' or black are the colours usually employed.
Black may be taken here as indicating metals and white as
colours. Furs, erynme and vair^ are also used ; but of these
ornamentation takes no account. Figs. 469 to 472 show the
principal divisions of shields. Fig. 469, checquy^ a shield divided
Fig. 469.
Fig. 470.
Fig. 471,
Fig. 472.
in chequers or small squares like a chessboard ; the number
varies. Fig. 470, quarterly^ the field being divided into four
quarters. Fig. 471, the Pale or a vertical strip set upright in the
middle of the shield and onethird of its breadth. Fig, 472
represents a band division. The upper part is the chiefs occu
pying onethird of the height, the fcss is the bar, horizontally
placed in the middle of the field.
Among the other divisions of the field must be reckoned the
chevron^ a A shaped strip and the cross, usually a Greek cross of
equal arms. When plain, this cross is in breadth onethird of
the shield ; but its varieties are manifold.
Diaper, Chequer, Spot, Powder.
ct diaper is a repeated pattern covering a given surface
intervals ; for example. Fig. 473, which gives the appear
an
encaustic tiling with no variation of pattern ; Fig. 474
imple of a chequer^ which consists of a repeat alternately
vviih. v<.v:ant spaces.
ORNAME/TATIQTxN.
209
Fig. 475 is 2iSpot pattern ; it must be observed that the spot
pattern has large regular vacant intervals, the chequer, a vacant
space equal to that of the repeat.
it
Fig. 475
<>
^
O <
Fig. 476.
la
EI
13
S
a
lara
HSia
H
Fig. 477.
Fig. 478.
Fig. 479.
Fig. 480.
Powder again differs from spot in point of scale, the unit of
powdering should be small and simple, . Powdering consists of
small and insignificant units of repetition and may be combined
with spot. For example, Fig. 476 is a combination of spot
and powdering.
Fig. 477 is a stripe and band pattern. Units of pattern when
arranged in narrow lines are called ' stripe,' and when wider,
'band.' For instance, Fig. 478 is the elementary stripe and
band, the simplest form of this decoration. Fig. 479 is a
O
2IO GEOMETRICAL' i5^RAWING AND DESIGN.
chequered band, with stripes arranged for diagonal decoration ;
Fig. 480 is a paneUing, derived from bands, and Fig. 481 a piece
of parquet flooring. Fig. 482 is an application of diaper or
chequer in more elaborate form. It represents an inlaid work
of independent, interlacing squares. In Fig. 483, crosses are
Fig. 4S1.
Fig. 482.
Fig. 4?3
Fig. 484.
arranged as a diaper for mosaic decoration. Fig. 484 is a chequer
formed by a combination of Greek fret and square foliage ; Fig.
485 is a more elaborated diaper or spot pattern on a circular
basis and constructed on a 45° net, which forms part of the
design. Fig. 486 is a diaper pattern formed in marble mosaic,
from San Vitale, Ravenna. Fig. 487, a ceiling panelling
ORNAMENTATION.
Fig. 4S5.
wm
^
m
u
ilL
.^
P^P^P
1
nr
^
i:
^
Fig. 486.
Fig. 487.
212 GEOMETRICAL DRAWING AND DESIGN.
conveying the impression of spot and powdering. It is taken
from a mediceval enamel in Cologne. Fig. 488 is a mosaic
%M'
^
((1^))
(TA ) (?•) T
Fig. 48E
flooring of simple construction and a charming intricacy.
Fig. 489 is one panel of ceiling decoration, in which only one
Fig. 489.
repeat is shown. Fig. 490, a scalework diaper, a simple
construction if based on a 45° net. Fig. 491, a very pleasing
combination of square and circular treatment suitable for a
mosaic flooring. Fig. 492 is a reproduction of a Byzantine
basrelief from the Cathedral of San Marco, Venice. It is
ORNAMENTATION.
Fig. 490.
Fig. 491
r
—
^
^^^^^
^
Fig. 492.
214
GEOMETRICAL DRAWING AND DESIGN.
a beautiful example of varied interlacing. The constructions
are extremely simple, and the efifect is due to the unexpected,
Fig. 493.
which piques the interest. Fig. 493 is an Egyptian ceiHng
decoration ; in reality it is a simple diaper and spot with spiral
■^^^^^"^
r
Fig. 494.
construction. Fig. 494 is the corner of a rectangular mosaic.
Fig. 495 is an interlacement band ornamenting a northern MS.
of the eighth or ninth century ; Fig. 496, a geometrical band
decoration in coloured marble from the wall of Mackworth
Church, Derbyshire. Fig. 497, a band of Moorish mosaic from
ORNAMENTATION.
215
Granada, which, it may be observed, is interchangeable, the
black and white spaces being exactly equivalent.
Fig. 495
Decoration based on geometrical construction is of infinite
variety, and a thorough familiarity with geometrical relations
will regulate genius and inspire the designer with a just view of
Fig. 496.
^^
Fig. 497.
those forms which will be pleasing and restful to the eye, as well
as satisfying to the natural demand for relief.
{For Exercises see p. 233.)
ARCHITECTURE.
CHAPTER XVIII.
Arch Forms and Tracery,
The semicircle was the first archform to appear in building.
In the earhest times (Nippur, B.C. 4000), such arches were
employed below the level of the ground. The tendency of the
round arch to sink when bearing any weight led these early
builders only to use it with that strong lateral support. The
Romans employed it largely ; as their bridges and aqueducts
standing to this day exemplify. The Pont du Gard, near
Nimes (B.C. 19), is about 160 ft. high and 880 ft. long ; built of
cJeryii circular
Fig. 498.
Fig. 499.
Fig. 500.
large stones without cement, it affords a striking example of the
stability of the semicircle.
The round arch may be semicircular, as Fig. 498 ; segmental,
as Fig. 499 ; "elliptic" or threecentred, as Fig. 500. It maybe
"stilted," as Fig. 501, the semicircle being continued in straight
ARCHITECTURE. 217
lines ; or a " horseshoe," as Fig. 502, the circle itself being
continued. This last is a Moorish feature, the semicircular a
Roman, and the stilted arch a Byzantine feature.
The Basilica, or Hall, originally a place of business, became
in the hands of the Roman builders a structure of heavy round
Fig. 501. Fig. 502.
arches and circular windows. This form was adopted for the
early Christian churches ; we have such a building at Brixworth,
Northants.
The Norman builders (1066 to 1190) followed up the semi
circular brick arches, building them in stone ; the heavy round
pillars with cushioned capitals, and the heavy arches with
geometrical ornamentation being the successors of the Roman
work.
The origin of the pointed arch has been much discussed ; an
example is found in Cairo, a horseshoe arch of the ninth
century ; in fact, at that time, this form was regarded by the
Moslems as their special emblem. It is met with in the
Crusaders' churches throughout the twelfth century. If in Eng
land it has been developed from Norman arcading, as is often
supposed, there is an example of the process on the towers of
Southwell Minster. There is a semicircular arcading ; then on
another face, semicircular arcades intersect ; the lancets thus
formed become windows in the next, and, finally, the arcading
disappears and lancetheaded windows are seen alone.
The Transitional Period, heavy Norman work with pointed
arches (about 1140 to 1200), is well exemplified in the arcading
of St. John's Church, Chester ; best, however, in the Choir of
Canterbury Cathedral.
The Early English, or Lancet Style (about 1190 to 1300), was
thus developed at the end of the twelfth century from the
circular or Romanesque ; the lancet windows (Fig. 503) and the
clustered pillars (Fig. 543) giving elegance and lightness.
Characteristic ornament was introduced, and a style grew up
which is essentially English.
2lJ
GEOMETRICAL DRAWING AND DESIGN.
At this time also Tracery begins to appear, apparently
developed as follows. Lancet windows are placed together ; in
Salisbury Cathedral, our great instance of a complete Early
English building, are seen combinations of two to even seven
lancet windows together. Again, three windows are often placed
Fig. 504.
together, the middle window being higher than the other two.
The three would be treated as one window, one arch moulding
including them all. Again, two lancet windows are placed close
together with a trefoil or quatrefoil (Fig. 512) above them, and
are treated as one window ; in Westminster Abbey, a small
triangle appears besides the small quatrefoil. This "wall
tracery " led to " plate tracery," where the wall is thinned to a
Fig. 506.
single piece of stone ; and plate tracery led to "bar tracery,"
where the wall between windows becomes a bar, and the
quatrefoil becomes a geometrical design in the arch.
ARCHITECTURE.
219
This leads to the Geometrical Period (about 1260 to 1320),
and, at this point, the study of tracery begins. The window is
divided, according to its size, by a number of vertical bars or
"mullions" ; the arch is equilateral (Fig. 504) and filled with
circles, trefoils and curved triangles in strictly geometrical
design. The " element " of the design (Fig. 509) is the skeleton
showing the centres of the circles of the tracery. The element
of some windows is nothing but a 45° network, with quatrefoils
as a unit of pattern. The element (Fig. 509) will form an
interesting exercise for the student to complete, and the ruined
window at the end of the book (p. 262) will be found a more
difficult problem of the same sort. In the Geometrical Period,
circular foils (Fig. 512) alone are used ; in the last quarter of the
thirteenth century, pointed foils (Fig. 513) are introduced.
The Decorated Period (about 1300 to 1375). The simplicity
of style, so far described, led, in the hands of ingenious designers,
to flowing curves, such as Figs. 511, 515, 518 suggest. This
flowing style is a purely English development ; it may be called
Flowing Decorated (about 131 5 to 1360). The "ogee" arch
(Fig. 507) and the pointed foiled arch are introduced ; also
Fig. 508.
bands of wavy foliage of a natural design. But this leads us
beyond the scope of Geometrical Design.
It seems, however, that the excess of ornament led to a return
of simpler forms in the last quarter of the fourteenth century,
and introduced the Perpendicular Style (about 1400 to 1545)
This deserves our attention because of its geometrical character.
GEOMETRICAL DRAWING AND DESIGN.
Straight lines are the feature of this style in place of the flowing
lines of the later Decorative Period, circular lines and circular
cusping. Horizontal transoms are introduced into the large
windows at this period, and become a decorative feature. The
I
1/
%
Fig. 510.
P^ig. 509.
Fig. 51
enclosing square which is used for construction in the element
509 becomes the outline of door and window. The fourcentred
arch is due to this period (Fig. 508).
This, again, is a style which is peculiarly English, and is
more appreciated nowadays. Perhaps it is too mechanical, and
Fig. 512.
Fig. 513
there may be too much repetition of ornament ; but to this
period belong the great East Anglian churches, and such
masterpieces as the Tower, Choir, and Lady Chapel of Gloucester
ARCHITECTURE.
221
Cathedral. It forms a grand close to the development of our
English Church Architecture.
Fig 517
Fig. 518.
Greek and Roman Mouldings.
Mouldings have been called the alphabet of architecture ;
they are the elements which determine and give expression
to the parts of a building.
The simple forms of moulding are :
The Roman ovolo, a quadrant (Fig. 519).
S
C
Fig. 519
Fig. 520.
Fig. 521.
The cavetto, or hollow (Fig. 520), which is the reverse of
the ovolo.
The torus, or halfround (Fig. 521).
222 GEOMETRICAL DRAWING AND DESIGN.
Other mouldings and cornices or other designs can be made
by arranging these with flat spaces, above, below, or between
them.
A fillet (Fig. 522) is a small flat face, and the torus when
small is called a bead or astragal (Fig. 523), which may be
Fig. 522. Fig. 523. Fig. 524.
incised, so as not to project from the flat surface. Several
parallel beads together are called reeding.
The cyma recta (Fig. 524) is formed by combining ovolo
and cavetto, the hollow being uppermost, and is suitable for
a cornice.
The cyma reversa, or ogee (Fig. 525), a similar combination,
with the hollow at the bottom, is suitable for a base
moulding.
The scotia is formed by two quadrants, as shown in Fig. 526.
Fig. 525
Fig. 526. Fig. 527
The Greek mouldings correspond with these, but their
section is not circular like the Roman. They^ are for the
most part constructed with conic sections, viz. ellipses, or
parabolas. But in all probability they were drawn in by hand.
Some construction lines for these mouldings are shown in the
figures, 527, ovolo; 528, cyma recta; 529, cyma reversa or
ogee ; 530, scotia.
ARCHITECTURE.
The Bird's Beak or Hawk's Beak moulding (Fig. 531) is
common in Greek Doric architecture. It is a cyma recta
surmounted by a heavy ovolo which casts a bold clear shadow
Fig. 529.
Fig. 530.
Fig 531
over the cyma. It is particularly interesting, because it dis
appears from architecture entirely after the best period of
Athenian art.
Greek architecture is distinguished by the grace and beauty
of its mouldings ; it has been remarked that their sections
are mostly elliptic. They are, however, not regular curves ;
they must be drawn, rules cannot be given for describing
them. Symmetry, proportion, and refinement are the charac
teristics of Greek ornament.
The mouldings of Roman origin are in general form the
same as the Grecian, but their contour is bolder and section
circular.
The ornamentation of Roman mouldings was no doubt
borrowed from Greece, but it is less restrained. Roman
architecture is overdone with ornamentation ; foliage and
various subjects in relief covering every moulding and surface.
Fig 532
Fig 533
One or two characteristic specimens of ornamentation may
be given. Fig. 532 is the egg and tongue or arrow on the
Greek echinus or ovolo ; and Fig. 533 the egg and dart on a
224
GEOMETRICAL DRAWING AND DESIGN.
Roman ovolo. A bead or astragal may be divided up,
as Fig. 534. Figs. 535 and 536 ^ \/^\/ — V^
are typical ornamentations of torus V TV /\ f\ J
moulding. Fig. 534
Fig. 535 Fig. 536.
Fig. 537 is a Greek leaf ornament for cyma reversa from
the Erechtheum at Athens, and Fig. 538 a Roman leaf ornament
for the same type of moulding.
Fig. 537 Fig. 538.
Gothic Mouldings.
Mouldings were developed contemporaneously with the other
features of Gothic architecture. In the Norman period, as might
be expected, these are square and circular in section.
The stock Norman moulding (Fig. 539) consists of a broad
hollow surmounted by a broad fillet, from which it is cut off by
a small sunk channel ; in fact, the hollow is set off by ' quirks ' or
returns.
Fig. 539
Fig. 540.
A plain round projection is frequently found with a narrow
fillet above it, the quirks being chamfered (Fig. 540). This
moulding is called a bowtel.
ARCHITECTURE.
225
There are very deep mouldings over the round Norman
doorways, consisting simply of squares and circles, at Iffley,
Oxford, for example (Fig. 541).
Fig. 541
Fig. 542.
The Early English architects developed this style of mould
ing, retaining circular forms and almost entirely eschewing ogee
or reversed curves. A single specimen only need be given,
part of a doorway, at Woodford, Northants (Fig. 542).
It may be noticed here that fillets are freely run down the
face of circular mouldings, that each circle is defined, and that
there is no returned curve. These may be taken as the simple
characteristics of Gothic mouldings.
Gothic Piers.
One architectural feature which should be mentioned is the
©development of the simple circular
and square pier of the Norman
style into the elaborate piers of the
Perpendicular period.
The massive Norman pier in the
hands of the early Enghsh builders
was lessened in size and had shafts
set round it, as Fig. 543, a specimen
Ofrom Salisbury Cathedral or the
north transept of Westminster
Abbey. The shafts increased in
Fig. 54^.
number and were incorporated in
the body of the pier, still preserving a circular contour.
p
zb
GEOMETRICAL DRAWING AND DESIGN.
The contour became of a lozenge plan in the style called
Decorated, and the number of pillars is much increased. The
shafts were arranged diamond wise, so many as would stand
close together, with only a fillet or small hollow between them.
i:Z\
Fig 544
Fig 545
Fig. 544 shows a Decorated pier at Dorchester, Oxfordshire,
and Fig. 545 a Perpendicular pier from Rushden, Northants.
{For Exercises see p. 235.)
MISCELLANEOUS EXERCISES
FROM EXAMINATION PAPERS OF THE
BOARD OF EDUCATION
ORTHOGRAPHIC PROJECTION (Ch. XIV.).
1. A plan and elevation are given of a l)uUress projecting from a
wall (Q. i). Draw a fresh elevation on a vertical plane which makes
an angle of 45° with the plane of the wall.
Q.I.
2. The end elevation is given (Q. 2) of a small cofter or caddy, the
length of which is to be 3". The lid has four sloping faces, which all
make the same angle (30°) with the horizontal. Draw the plan of the
lid.
228
GEOMETRICAL DRAWING AND DESIGN.
3. The plan is given (Q. 3) of an octagonal tray or dish, the height
of which is ". Make an elevation on the given xy. Only the visible
lines need be shown in elevation, and the thickness of the material is
to be neglected.
4. The diagram (Q. 4) represents a doorway in a wall, the door
being shown opened at an angle of 45° with the surface of the wall.
Draw an elevation of the door when closed, t'.e. showing the true form
of the panels. On/j' the door need be drawn, not the surrounding
mouldings. Your construction 'uusi be shown.
Q5
5, The plan is given (Q. 5) of a square slab, a5 being one edge of a
square base. Draw an elevation on a vertical plane at right angles to
ad. Show in plan the section made by a horizontal plane containing
ad.
Q. 6.
I. — .....J
Q.7.
6. The plan and end elevation are given (Q. 6) of a simple hut.
Draw an elevation on a vertical plane w^hich makes an angle of 30"
with the long walls of the hut. (N.B. — Only the visible lines need be
shown. )
7. The diagram (Q. 7) shows an elevation and section of an opening
in a wall. Make a second elevation when the face of the wall makes
an angle of 45° with the vertical plane of projection.
MISCELLANEOUS EXERCISES.
229
8. The diagram (Q. 8) shows an elevation of a square slab, Ji B
being one side of a square face, 3" long. Draw the plan of the slab,
and show a cylindrical hole of 2" diam. pierced through its centre.
9. The plan is given (Q. 9) of a square prism, of which AB represents
a square face, 3" each side. Determine the elevation of the prism on
the given xy, and add the elevation of a circular hole of 2" diameter
piercing the centre of the prism.
10, The diagram (Q. 10) shows a perspective sketch of part of a
buttress. Make an approximate sketch plan and a side elevation.
SECTIONS OF SOLIDS (Ch. XV.).
1. Make an approximate sketch plan, and also a sectional elevation,
of the mortar of which a perspective sketch is given (Q. i;, assuming
that the inside form is a hemisphere. Show clearly any construction
you would suggest. The size ot your drawing should be about 3 times
that of the diagram.
230 GEOMETRICAL DRAWING AND DESIGN.
2. A right cylinder of 2" diameter is cut by a plane, making an
angle of 30° with the axis of the cylinder. Show the true form of the
section.
3. The plan is given (Q. 3) of a right cone (diameter of base, 2^"),
of which y is the vertex. Make an elevation on the given xy, and show
the section by a vertical plane parallel to the base and i" from it.
DESIGN.
CONSTRUCTION LINES AND UNITS, pp. 184195.
1. Sketch four illustrations of ornament formed by circle?, and
explain the object they fulfil in certain cases.
2. Draw clearly, with instruments or freehand, the geometrical basis
on which the given "diaper" pattern (Q. 2) is constructed. Plan the
scale of your diagram to show two "repeats" of the pattern in a
width of 3^". Show 5 or 6 repeats in all. (Only sufficient of the
ornamental detail need be sketched to indicate its position.)
Q. 2. Q. 3
3. Sketch, with instruments or freehand, 07ie unit of the given
diaper (Q. 3), showing clearly your method of setting out its details.
Make your drawing about three times as large as the diagram. [The
height and width of the figure are equal.]
4. Draw, freehand or with instruments, a system of construction
lines on which the given system of quatrefoils (Q. 4) can be built up.
Show how you would determine the centres of the arcs, and the points
of contact. Three or four repeats of the unit should be indicated about
twice the size of the diagram.
MISCELLANEOUS EXERCISES.
5. Show clearly any geometrical construction you would think useful
in setting out the given "Tudor rose" (Q. 5) about four times the
dimensions of the diagram. You need only sketch so much of the
flower as is needed to illustrate your construction.
cj. 4. Q. 5
6. Draw, with instruments or freehand, the system of construction,
lines on which you would build up the given repeating pattern (Q. 6).
Show also one complete unit of the repeat.
Q. 6.
Q. 7.
7. Show clearly how you would set about drawing the given pattern
(Q. 7). Mark what you consider the unit, and indicate any construc
tions you think needful. Only two complete adjacent units need be
shown, about four times the scale of the diagram.
8. Any triangle can be repeated so as to cover a space without leaving
interspaces. Show how this can be done, and sketch four other shapes,
rectilinear or curved, which will repeat in a similar way.
9. A floor has to be covered with tiles which are squares and regular
octagons in shape. Sketch the pattern so formed, showing clearly how
you would set it out, and marking the unit of repeat.
232
GEOMETRICAL DRAWING AND DESIGN.
SPACING OF SURFACES, pp. 196201.
1. Draw a Greek fret without keys, but with a border at top and
bottom ; the fret, borders and spaces, to be " wide. Also another fret
with a tee and a border top and bottom ; borders, spaces and fret each
\" wide. Explain the principles of the ornamentation and the surfaces
to which they can be applied.
2. Draw three sorts of frets, and another fret on the slant, and an
instance of frets alternating with ornamented panels, each i" high.
Explain the principles and the surfaces to which they are properly
applied.
3. Draw a band ornamented with square panels filled in with some
usual Greek ornament, divided from one another by Greek keyed frets,
meeting at centre line of panel, with border top and bottom : height
exclusive of border i", length 3", to contain three panels. Explain the
principles of the ornamentation.
Q.5.
4. Make a drawing of one unit of the given border (Q. 4), increasing
the length of the "unit" to 2\", and the other dimensions in proportion.
Indicate the method by which the figure should be constructed.
5. Make an enlarged copy of one unit of repeat of the given border
(Q. 5), the height of your drawing being increased to iA^"and the length
in proportion. Show a construction for obtaining the divisions of the
circle.
6. The diagram (Q. 6) shows half a circular plate. Indicate a
method by which the leading wave line of the ornament could be made
up of arcs of circles of equal radius.
Your drawing should be three times the diagram.
MISCELLANEOUS EXERCISES.
233
7. Show the construction you would employ in setting out the given
border (Q, 7).
Not more than three repeats should be drawn.
Q. 7.
DEFINED AREAS, pp. 202206.
1. Show the construction lines upon which the ornament in the
square panel (Q. i) has been designed.
2. The figure (Q. 2) shows one quarter of the decoration of a circular
plaque. Complete the circle and set out the panels. The position only
of the freehand ornament need be indicated.
Q.I.
Q. 2.
3. Show how you would proceed to modify the given figure (Q. 3) so
as to make the central panel a regular octagon, the width of the four
side panels remaining unchanged.
4. What is the geometrical basis of the given design (Q. 4) for chip
carving? (N.B. — Show only the main lines. Do not attempt to copy
the figure completely.)
234 GEOMETRICAL DRAWING AND DESIGN.
5. Draw a coffered ceiling, 6" square, with beams round the outside,
with a circular panel in the middle, and four angle or spandrel panels,
the spacing to be in harmonic proportion. Ornament the coffers and
the soffits of the beams if you can, and explain the principles of the
ornamentation.
ORNAMENTATION, pp. 207215.
1. Draw, with instruments or freehand, two different arrangements
by which the two given ornaments (Q. i) may be used, alternating with
one another, so as to form a ' •' diaper " pattern.
(The ornaments may be roughly sketched simply to indicate their
position. )
2. Show how, by repeating and reversing the given lines (Q. 2), an
*' all over " pattern may be obtained. Indicate the lines of construction.
Q I. Q. 2. Q. 3. Q. 5.
3. Sketch, with instruments or freehand, an "all over" diaper
pattern formed by repeating the given unit (Q. 3). Show nine repeats
of the unit, with the leading lines of the construction. Make each unit
about three times the size of that in the diagram. (The height and
width of the figure are equal. )
4. The outline of the diaper pattern formed by placing repeats of the
given figure (Q. 3) in contact with one another is made up of semicircles.
Draw at least four repeats of the outline so as to show clearly where
centres and points of contact of the semicircles occur. Make each unit
about five times the size of that in the diagram.
5. The diagram (Q. 5) represents a stencilled ornament which it is
desired to repeat so as to form a "diaper" pattern. Sketch two ways
in which this can be done, the repeats of the unit being placed adjacent
to one another. Show four repeats in each case about twice the size of
the diagram.
6. Indicate clearly a geometrical basis for the given repeating pattern
(Q. 6), and show what you consider to be the unit. (The freehand
ornament should only be shown once.)
MISCELLANEOUS EXERCISES.
235
7. It is desired to restore the complete circular ornament of which a
fragment is shown (Q. 7). How would you do this?
The restorations of the dark portions should be disregarded altogether
in your drawing.
g. 8.
8. Indicate any geometrical construction you think desirable in setting
out the given pattern (Q. 8).
Show clearly what you consider the unit of the pattern. (N.B. — Do
not try to copy all the details ; only show enough to make your meaning
clear.)
9. Draw, with instruments, specimens of scale work (imbrication),
showing two scales and a half to each in length, each scale to be %"
wide. Give specimens of scales formed of half circles, oblongs with
rounded ends, outlines of leaves, leaves with an outer margin and ribs,
and one whose leaves are double ogees with the point of the leaf turned
up, and with ribs, and one of trefoils with margin and filled with
ornament. State to what surfaces they can be properly applied.
236
GEOMETRICAL DRAWING AND DESIGN.
ARCHITECTURE, pp. 216225.
1. Draw the given diagram of window tracery (Q. i), using the
figured dimensions. The arch is " equilateral."
^■f'^S
Q. 2.
Q3
2. Draw the given outline of window tracery (Q. 2), using the
figured dimensions. The arch is "equilateral" and all the arcs are of
equal radius.
3. Draw the "cyma recta" moulding shown (Q. 3), adhering to
the given dimensions. The curve is composed of two quarter circles
of equal radii, tangential to one another and to the lines AB and CD
respectively.
4. Draw the "scotia" moulding shown (Q. 4). The curve is
made up of two quartercircles of i" and ^" radius respectively.
Q. 5
O. 6.
5. Draw the
dimensions.
"rosette" shown (Q. 5), according to the figured
6. Draw the "ogee" arch shown (Q. 6) to a scale of 2' to i".
The arcs are all of 2' radius. The methods of finding the centres and
points of contact must be clearly shown,
7. Draw the moulding shown (Q. 7), adhering strictly to the
figured dimensions. The arc of V radius is a quadrant.
8. Diaw the " cyma recta " moulding shown in the diagram
(Q. 8), using the figured dimensions. The curve is composed of two
equal tangential arcs each of f" radius.
MISCELLANEOUS EXERCISES.
237
9. Copy the cornice given (Q. 9), increasing the total height to 2"
and the other measurements in proportion. "\'ou may draw the "cyma"
moulding by any geometrical construction that seems to you suitable.
Q. 7.
■■m'>
1
/^
r '^
J
\
Q. 8.
^
Y
Q. 9.
10. How would you draw the leading lines of the window tracery
given (Q. 10)? (N. B. — Do not try to copy the whole figure, but only
clearly indicate your method.)
11. Show how to set out the given figure (Q. 11).
12. Make an approximate sketch plan or elevation (but not both) of
the given column base (Q. 12).
EXAMINATION PAPERS IN GEOMETRICAL
DRAWING.
BOARD OF EDUCATION.
GENERAL INSTRUCTIONS TO CANDIDATES.
You may not attempt more than five questions, of which three only
may be chosen from Section A, and two only from Section B. But no
award will be made to a Candidate unless he qualifies in both sections.
All your drawings must be made on the single sheet of drawing paper
supplied, for no second sheet will b& allowed. You may use both sides
of the paper.
None of the drawings need be inked in.
Put the number of the question close to your workings of problems,
in large distinct figures.
The value attached to each question is shown in brackets after the
question.
A single accent (') signifies_/^^/ ; a double accent (") inches.
Questions marked (*) have accompanying diagrams.
Your name may be written only upon the numbered slip attached to
your drawing paper.
I.
SECTION A.
In this section you may attempt three questions only,
SHOWING your knowledge BY THE USE OF INSTRUMENTS.
The constructions must therefore be strictly geometrical, and not the
result of calculation or trial.
All lines used in the constructions niust be clearly shown.
Set squares may be used wherever convenient. Lines may be bisected
by trial.
EXAMINATION PAPERS.
239
1. Draw a diagonal scale yV of full size, by which feet, inches, and
eighths of an inch may be measured up to 5'.
By means of this scale construct a triangle having its altitude 2' 2J",
one side 2' 5I", and base 2' o". Write down the length of the third
side. (20)
*2. Draw the given figure by inscribing seven equal squares within
a circle of i" radius. (20)
Q.3
Q. 6.
Q. 7.
*3. Copy the given guilloche ornament, making the radii of the
circles ", ^", ", and i" respectively. (16)
4. Describe two circles of f" radius touching each other, and one of
i" radius touching the first two. Then describe a fourth circle touching
all these three. All constructions must be clearly shown, and all
contacts indicated. (20)
5. A quadrilateral ABCD is to be described about a circle of "
radius. AB = 2%", AD=2h,", and the angle BAD is 42°. Draw the
figure, show all points of contact, and write down in degrees the angle
ADC. Then draw a similar quadrilateral having the radius of its
inscribed circle i". (18)
240 GEOMETRICAL DRAWING AND DESIGN.
*6. An elevation is given of part of an octagonal pillar with square
base. Draw its plan, and an elevation on a vertical plane which makes
30° with one of the vertical faces of the base. (20)
*7. The diagram shows the elevation of a solid composed of a
cylinder capped by a portion of a sphere. Draw the true form of the
section made by the plane indicated by the dotted line. (20)
SECTION B.
In this section you may attempt two questions only.
All freehand work employed in this section must be neat and
careful, and its intention must be made quite clear.
All constructions must be clearly shown.
*8. Redraw the given pattern, with instruments, altering the pro
portions so as to make the octagons regular and of " side. (20)
Q. 8.
*9. Show the geometrical framework on which the given pattern is
based. Show clearly what you consider the unit of repeat, and draw
four repeats, sketching only enough of the pattern to make your
intention plain. (18)
*10. Draw the geometrical constructions you think necessary in
setting out the plate, making your drawing double the dimensions of
the print. It will suffice if one quarter of the design is clearly shown.
(20)
"■■"11. Draw, using instruments, one of the cusped archforms, pre
serving, as accurately as you can, the proportions of arch and cusping.
Make your drawing twice the dimensions of the print. (20)
EXAMINATION PAPERS.
241
*12. Make an approximate sketch plan and front and side elevations
of the flowerholder, using instruments where you think advisable.
Arrange your drawings so that one is projected from another. (20)
Q. 9.
242 GEOMETRICAL DRAWING AND DESIGN.
Q. II.
EXAMINATION PAPERS. 243
Q. 12.
244
GEOMETRICAL DRAWING AND DESIGN.
II.
SECTION A.
1. A drawing made to a scale of ^ of full size has to be redrawn so
that the dimensions shall be enlarged by onefifth. Make a scale for the
new drawing, showing feet (up to 3'), inches, and (diagonally) eighths of
an inch.
Figure the scale properly, and show by two small marks on it how
you would take off a distance of i' 4§". (22)
g. 2. Q. 5.
■^2, Copy the given figure, making the diameter of the outer circle
38". Show how to determine all the points of contact between circles
and straight lines. {20)
Q. 6.
Q. 7.
3. Two points, A and B, are i" apart. Find a third point, C, I "6"
from A and 2" from B. With centre A and radius AB describe a circle.
Describe a second circle touching the first at B, and passing through C.
Describe a third circle of i" radius touching the first two (but not at B).
(18)
4. The foci of an ellipse are 3" apart. A point Z>, on the curve, is i"
from one focus and 3" from the other. Draw the ellipse and a normal
to it at D. (16)
*5. The point of intersection of the two given lines being inaccessible,
draw through the point Z' (i) a line which would pass through the point
EXAMINATION PAPERS.
245
of intersection of the two given lines, and (2) a line making equal angles
with the two given lines. (16)
*6. The diagram shows the plan of a right square pyramid. Draw an
elevation. (18)
■^7. The elevation is given of an "elbow" formed by two pieces of
cylindrical piping. Draw the plan, and also the true form of the inter
section of the pipes. The thickness of the material may be neglected.
(22)
SECTION B.
*8. Show how you would set out a geometrical framework for the
given pattern, so as to exhibit a number of repeats. Only enough of the
246
GEOMETRICAL DRAWING AND DESIGN.
ornament should be sketched to show quite plainly what you consider
the unit of repeat. (18)
9. Regular pentagons will not by themselves cover a surface. Draw
any form which in combination with regular pentagons would serve this
purpose. Make a diagram of the pattern formed, and show clearly what
you would use in practice as the unit of repeat. .' (lb)
*10. Show what geometrical aids you would employ in drawing the
square panel given. Make your drawing about the size of the figure.
(18)
■^11. What geometrical means would you use in setting out the fan
shown? Make your drawing about the size of the figure. (16)
*12. Draw a plan of the given table, showing clearly any construc
tions you would use. (18)
EXAMINATION PAPERS.
247
A=^5i
ift?^
ii;??:«; irrt'V '
c .viv'ta?..
248
GEOMETRICAL DRAWING AND DESIGN.
III.
SECTION A.
1. Make 2. plain scale, to show feet and inches up to 5 feet, on which
a distance of 3' 6" is represented by 4I". Finish and figure the scale
neatly and carefully.
Draw to this scale an oblong 3' 3" x 2' 5", and in the centre of it
place a second oblong of the same shape but having its longer sides
2' 9". Measure the breadth of this smaller oblong to the nearest half
inch. (20)
Q. 2.
Q. 3.
Q. 4.
Q. 6.
Q. 7.
*2. Complete half \h^ given figure, which is made up of regular
pentagons. Make the radius of your enclosing circle 2". Show clearly
any construction you employ. (20)
*3. Copy the given figure, making the radius of the outer circle if,
and that of the inner one in proportion. (18)
*4. Draw the given figure. The curves are to be composed of arcs
of circles of 05" and 15" radii. (20)
EXAMINATION PAPERS.
249
5. Two straight lines, AB and CD, are 3" and 35" long respectively.
A is 15" from C, while B is i" from D and 35" from C. Describe two
circles each touching both AB and CD, one passing through A, the
other through ^. (16)
*6. The outline of a "scotia" moulding is shown. If two lengths
of this moulding are "mitred" together at rightangles, show the true
form of the cut surface of the mitre. (18)
*7. A lampshade, in the form of a truncated regular hexagonal
pyramid, is made of six pieces of card, of the exact shape and twice the
size shown. Draw its plan and elevation in any position. (20)
SECTION B.
*8. Show, using instruments, how you would set out the geometrical
framework of the given openwork panel. None of the "cusping"
need be drawn. (20)
Q. 8.
*9. Draw neatly with instruments the framework or "net" of the
given pattern. Show clearly what you intend to be the unit of repeat,
and finish not less than four of these so as to show repeats both in
width and height. (2o)
250
GEOMETRICxVL DRAWING AND DESIGN.
*10, Set out, as nearly as you can, the construction needed for the
geometrical part of the ornament round the semicircular doorhead of
which about half is shown. Only one unit of each ornament need be
completed. Make your drawing twice as large as the diagram. (i8)
Q. 9.
EXAMINATION PAPERS.
251
"^11. Redraw the given figure, altering the proportions of the parts so
that the sides of the four corner panels shall be exactly half those of the
centre panel. Make the side of the outside square 3", and the margins
throughout 02" wide. (16)
D
1 I
D
D
1 1
D
Q. II.
'^12. Draw an approximate sketchplan, with front and side eleva
tions, of the given steps. Arrange your drawings to show how one is
projected from the other. (18)
252
GEOMETRICAL DRAWING AND DESIGN.
IV.
SECTION A.
1. Six feet are represented on a drawing by one inch. Make a scale
for the drawing by which single feet can be measured up to 40', and show
inches diagonally.
Figure the scale properly, and show by two small marks on it how you
would take off a distance of 20' 8". (22)
■^2. Make a copy of the diagram, using arcs of I J" and " radii only.
Show clearly how all points of contact are obtained. (18)
*3. Copy the diagram, making the radius of the outer circle if".
(20)
■^4. The diagram shows a symmetrical figure composed of straight
lines and five semicircles of equal radii. Draw a similar figure having
a total height of 3^". (22)
Q. 4.
Q. 6.
5. Construct a regular nonagon of if" side. Describe a circle touch
ing all the sides of the nonagon. Within the circle inscribe a regular
nonagon having its sides parallel to those of the first one.
If you employ a protractor for measuring an angle, this must be
clearly shown and the number of degrees stated. (16)
EXAMINATION PAPERS.
253
*6. The diagram shows the elevation of a short prism, or slab, the
bases of which are eqiiilataal triangles. Draw the plan, and write
down the angle which the bases make with the vertical plane of pro
jection. (18)
7. A right cone, diameter of base 3I", height 2^", has the plane of its
base inclined at 45° to the horizontal plane. Draw the plan of the cone,
and of its section by a plane parallel to the base and i^" from the vertex.
(20)
SECTION B.
*8. Draw a geometrical framework on which the given diaper pattern
may be constructed. You are advised to draw the framework geometri
Q.8.
254
GEOMETRICAL DRAWING AND DESIGN.
cally, and to sketch, freehand, just enough of the pattern to illustrate
your meaning. Show very clearly what you consider to be the unit of
repeat. (i8)
*9. The diagram shows a pattern composed
entirely of semicircles. Show two other ways
of arranging semicircles so as to produce a
repeating pattern, with the necessary construc
tions for determining centres. (i6)
•^10. Show what geometrical help you would
use in setting out the plate shown in the
diagram. Assume that AB, the diameter of
the octagon, is 5". (16)
*11. Indicate what geometrical construc
tions you would employ in setting out the circular window shown in
the diagram. The "cusping" need not be shown. (18)
Q. 9.
Q. 10.
■^12. Sketch appro.ximate front and side elevations of the chair shown
in the diagram. (ij^j
EXAMINATION PAPERS. 255
Paper IV. Q. 11.
256 GEOMETRICAL DRAWING AND DESIGN.
Paper IV. Q. 12.
EXAMINATION PAPERS.
257
SECTION A.
*1. The given line AB represents a length of 15 centimetres. Con
struct a scale by which decimetres, centimetres, and millimetres can be
measured up to 3 decimetres. Figure the scale properly, and show by
two small marks on it how to take off on it a distance of 263 milli
metres.
By means of the scale draw a ciicle of 125 millimetres' radius, and in
it place a chord 189 millimetres long. Write down in millimetres the
distance of this chord from the centre of the circle.
(i metre = 10 decimetres, or 100 centimetres, or looo millimetres.)
(20)
B
2. Construct a regular decagon or llgure of 10 sides, each side being
1" long. Within it mscnhQ Jive equal circles, each touching two of the
others and one side of the decagon.
(N.B. — If a protractor is usecl for measuring an angle, such use must
be clearly shown.) (20)
*3. The curve of the given " scotia " moulding is made up of two
quadrants or quartercircles. Draw the figure from the given dimensions.
(N.B.— The diagram is not drawn to scale.) (16)
Z'4
Q. 3 Q 4
*4. The diagram shows the leading lines of a window composed of
three semicircularheaded " lights " included under a threecentred arch.
The side lights are to be each 3' wide, while the middle light is to be
2' wide and to have the centre of its semicircle 2' above those of the
other two. Parts of the side semicircles also form part of the enclosing
arch. Draw the figure to a scale of 2' to 1". Show clearly how you
obtain all points of contact. (20]
(N.B. — The diagram is not drawn to scale.)
258
GEOMETRICAL DRAWING AND DESIGN.
6. Two fixed lines AB and CD, of indefinite length, cross one another
at an angle of 70°. A third line EF, 3" long, is movable, so that the
end E travels along the line AB while the end F travels along the line
CD. Draw the complete curve traced by the middle point of EF.
(It will be sufficient to find some 12 to 16 points on the curve.)
(18)
■''6. The diagram gives the front elevation of a regular fivepointed
star, which is cut out of material 'f thick. Draw the plan of the star,
and also a second elevation on a vertical plane inclined at 60° to that of
the given elevation. (20)
Q. 6.
*7. An elevation is given of a sphere upon which rests a conical
lampshade. Draw the plan of the shade. (20)
Q. 7.
EXAMINATION PAPERS.
259
SECTION B.
*8. Draw with instruments the main geometrical construction lines
you would employ in setting out the given pattern. Show plainly what
you consider the practical or working unit of repeat. (20)
9. Show how you would arrange a number of circular discs of "
radius as a diaper pattern —
(i) When each disc touches four others,
(ii) When each disc touches six others,
(iii) When each disc touches three others.
Draw the necessary constructional framework in each case, and show
clearly what you consider the unit of repeat in each pattern. (20 marks).
26o GEOMETRICAL DRAWING AND DESIGN.
_ *10. Set out carefully the lines of one quarter of the given book
binding, making your drawing to twice the scale of the diagram.
■•■11. Make a modified version of the border of interlacing circles
which the radii shall be ^", g", §", and i" respectively. Show at 'l(
two repeats of the unit.
EXAMINATION TAPERS.
261
*12. Sketch approximately, in outline, the plan and the front and
side elevations of the workbox and its lid shown in the diagram,
omitting all merely ornamental details. Arrange your drawings so as
to show how one is projected from another. (20)
Q, .2.
262 GEOMETRICAL DRAWING AND DESIGN.
Complete the design of this ruintd window, following the indicj
of every fragment \\ hich remains.
Date Due i
'^^(.Y 2 ^ S?
QP ^
DEC 2?
1993
X
Library Bureau Cat. No. 1137
WiELlS BINDERV
ALTHAM, MASS.
JAN. 1948
NC715.S7
^ 3, 5002 00047 5942
Spanton, J. Humphrey ^v^r«_
Geometrical drawing and design.
SCffi